nForum - Discussion Feed (functor) 2023-09-24T23:56:34+00:00 https://nforum.ncatlab.org/ Lussumo Vanilla & Feed Publisher Urs comments on "functor" (109690) https://nforum.ncatlab.org/discussion/332/?Focus=109690#Comment_109690 2023-05-20T08:47:37+00:00 2023-09-24T23:56:32+00:00 Urs https://nforum.ncatlab.org/account/4/ added pointer to: Saunders MacLane, §I.3 of: Categories for the Working Mathematician, Graduate Texts in Mathematics 5 Springer (1971, second ed. 1997) ...

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nLab edit announcer comments on "functor" (99662) https://nforum.ncatlab.org/discussion/332/?Focus=99662#Comment_99662 2022-06-07T16:21:23+00:00 2023-09-24T23:56:33+00:00 nLab edit announcer https://nforum.ncatlab.org/account/1691/ adding text from HoTT wiki Anonymous diff, v65, current

Anonymous

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Urs comments on "functor" (93865) https://nforum.ncatlab.org/discussion/332/?Focus=93865#Comment_93865 2021-07-12T17:02:47+00:00 2023-09-24T23:56:33+00:00 Urs https://nforum.ncatlab.org/account/4/ Wanted to point to the component function F x,y&colon;C(x,y)&rightarrow;D(F(x),F(y))F_{x,y} \;\colon\; C(x,y) \xrightarrow{\;\;} D(F(x), F(y)) of a functor, but it wasn’t mentioned here. ...

Wanted to point to the component function $F_{x,y} \;\colon\; C(x,y) \xrightarrow{\;\;} D(F(x), F(y))$ of a functor, but it wasn’t mentioned here. Have added it now to the first lines of the Idea-section, making some small adjustments in the course.

This ancient entry would need a bit of a clean-up…

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Dmitri Pavlov comments on "functor" (87904) https://nforum.ncatlab.org/discussion/332/?Focus=87904#Comment_87904 2020-11-21T01:18:48+00:00 2023-09-24T23:56:33+00:00 Dmitri Pavlov https://nforum.ncatlab.org/account/356/ Added a reference to Joyal’s CatLab. diff, v63, current

Added a reference to Joyal’s CatLab.

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Mike Shulman comments on "functor" (75900) https://nforum.ncatlab.org/discussion/332/?Focus=75900#Comment_75900 2019-02-05T20:22:28+00:00 2023-09-24T23:56:33+00:00 Mike Shulman https://nforum.ncatlab.org/account/3/ A homomorphism (between structures with identities, like monoids or groups) always preserves identities.

A homomorphism (between structures with identities, like monoids or groups) always preserves identities.

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Ali Caglayan comments on "functor" (75899) https://nforum.ncatlab.org/discussion/332/?Focus=75899#Comment_75899 2019-02-05T17:58:00+00:00 2023-09-24T23:56:33+00:00 Ali Caglayan https://nforum.ncatlab.org/account/1731/ Also the “categorification” should be of the notion of “identity preserving homomorphisms”?

Also the “categorification” should be of the notion of “identity preserving homomorphisms”?

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Mike Shulman comments on "functor" (75897) https://nforum.ncatlab.org/discussion/332/?Focus=75897#Comment_75897 2019-02-05T17:02:26+00:00 2023-09-24T23:56:33+00:00 Mike Shulman https://nforum.ncatlab.org/account/3/ In fact, I would say that both preservation of identities and composition are “the functoriality condition”. diff, v58, current

In fact, I would say that both preservation of identities and composition are “the functoriality condition”.

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nLab edit announcer comments on "functor" (75892) https://nforum.ncatlab.org/discussion/332/?Focus=75892#Comment_75892 2019-02-05T15:30:10+00:00 2023-09-24T23:56:33+00:00 nLab edit announcer https://nforum.ncatlab.org/account/1691/ Added preservation of identity to ideas section. It is misleading to think that a functor does this automatically. Take the terminal category 1\mathbf{1} and an idempotent monoid DD. A functor ...

Added preservation of identity to ideas section. It is misleading to think that a functor does this automatically. Take the terminal category $\mathbf{1}$ and an idempotent monoid $D$. A functor $F : \mathbf{1} \to D$ has the following: $F(f) = F(f \circ 1_{\mathbf{1}}) = F(f) \circ F(1_{\mathbf{1}})$ but if $F$ does not preserve identity then $F(1)$ may be the idempotent. I think when these are groupoids, preservation of identity can be dropped but idempotent categories require the restriction.

Anonymous

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Eric comments on "functor" (7478) https://nforum.ncatlab.org/discussion/332/?Focus=7478#Comment_7478 2010-04-03T20:31:06+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ Nice! Thanks Mike.If no one beats me to it, I'll add your suggestions to the nLab. Edit: And thanks Todd! Yes. Thanks so much. That all makes perfect sense. All my fears are gone :)

Nice! Thanks Mike.If no one beats me to it, I'll add your suggestions to the nLab.

Edit: And thanks Todd! Yes. Thanks so much. That all makes perfect sense. All my fears are gone :)

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Todd_Trimble comments on "functor" (7477) https://nforum.ncatlab.org/discussion/332/?Focus=7477#Comment_7477 2010-04-03T20:30:42+00:00 2023-09-24T23:56:33+00:00 Todd_Trimble https://nforum.ncatlab.org/account/24/ This comment is invalid XHTML+MathML+SVG; displaying source. &lt;div&gt; &lt;p&gt;[Edit: I&#039;m covering much the same ground as Mike just ... This comment is invalid XHTML+MathML+SVG; displaying source. <div> <p>[Edit: I'm covering much the same ground as Mike just did]</p> <blockquote> A part of me still wishes we could define a diagram as follows ... </blockquote> <p>Again, I personally don't mind our doing that.</p> <blockquote> I like this because diagrams are almost always (I can't think of a counter example) functors from free categories for some graph. </blockquote> <p>A counterexample would be a functor whose domain has just one object and exactly two morphisms. The non-identity morphism could be either idempotent or involutive. This category is not a free category.</p> <p>However, in the context of speaking of commutative diagrams, this sort of thing is arguably a little bit artificial.</p> <blockquote> In fact, that last sentence could probably be improved, because I'm willing to bet that any preorder is a quiver for some graph. </blockquote> <p>Sorry, not true. The commuting square, i.e., the poset consisting of subsets of ${0, 1}$ ordered by inclusion, is not a quiver (free category). Do you see why?</p> <p>(In fact, here's a little factoid I rather enjoy: if the free category on a directed graph has a terminal object, then the graph is a tree.)</p> <blockquote> I interpreted Mike's response to mean Definition #1 and Definition #2 were the same, but all Mike really said was #2 includes #1. For them to be the same, #1 would also have to include #2. I'm afraid that might not be the case. I'm afraid #2 might be stronger than #1. I'm not sure, but I think with #2, we WOULD have the image always being a subcategory. What do you think? </blockquote> <p>The two statements are equivalent. Definition # 1 is the standard definition of functor. It implies definition #2, because functors preserve commuting diagrams.</p> <p>Definition # 2 could be considered an axiom scheme, since you are quantifying over all diagrams in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align: -20%;" class="tex" alt="C"/>. Probably you see definition #1 more often because there are only two axioms.</p> <p>The image of a category <img src="/extensions/vLaTeX/cache/latex_dc3fcd71eea8cc9d58dab8ea82eae894.png" title="F(C)" style="vertical-align: -20%;" class="tex" alt="F(C)"/> not necessarily being a subcategory has nothing to do with preservation of commuting diagrams. It has to do with the image not necessarily being closed under compositions. I gave an example a while back.</p> </div>]]> Mike Shulman comments on "functor" (7475) https://nforum.ncatlab.org/discussion/332/?Focus=7475#Comment_7475 2010-04-03T20:03:01+00:00 2023-09-24T23:56:33+00:00 Mike Shulman https://nforum.ncatlab.org/account/3/ They are the same. If is a functor in the usual sense (i.e. definition #1), and is a commuting diagram factoring through a preorder , then since functor composition (in the usual sense) is ...

They are the same. If $G:C\to D$ is a functor in the usual sense (i.e. definition #1), and $Free(G) \to P \to C$ is a commuting diagram factoring through a preorder $P$, then since functor composition (in the usual sense) is associative, the induced diagram $Free(G) \to C\to D$ also factors through $P$.

Diagrams in the sense of "diagrams-we-draw-on-a-page" are basically always functors from a free category or equivalently graph maps. I think the point is that there is a different meaning of "diagram" which refers to an arbitrary functor. It is fairly common, for instance, to talk about $D^C$ as "the category of diagrams in $D$ of shape $C$" for arbitrary categories $C$ and $D$.

Not every preorder is free on a graph, in fact very few are. For instance, the commuting square is not.

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Eric comments on "functor" (7474) https://nforum.ncatlab.org/discussion/332/?Focus=7474#Comment_7474 2010-04-03T19:19:40+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ This comment is invalid XML; displaying source. &lt;p&gt;I should go through these 200+ comments and pick out pearls of wisdom and add them to the nLab. That is a small project and, time ... This comment is invalid XML; displaying source. <p>I should go through these 200+ comments and pick out pearls of wisdom and add them to the nLab. That is a small project and, time permitting, I will do my best.</p> <p>But for now, I thought I would summarize some of the key mental stumbling blocks that I think point to ways we can improve the nLab to help future "scientists and engineers" avoid the same problems I had.</p> <p>First, although this probably sounds silly, one thing that really hung me up was thinking that the image <img src="/extensions/vLaTeX/cache/latex_dc3fcd71eea8cc9d58dab8ea82eae894.png" title="F(C)" style="vertical-align:-20%;" class="tex" alt="F(C)" /> was a subcategory of <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" />. This one factoid caused me immeasurable mental duress. This should be stated clearly somewhere on <a href="https://ncatlab.org/nlab/show/functor">functor</a> I think. Because of this, I thought every diagram was a category. This should also be repeated on <a href="https://ncatlab.org/nlab/show/diagram">diagram</a>, i.e. a diagram is not necessarily a subcategory.</p> <p>For instance, since a diagram does not need to be a category, we can have diagrams like this</p> <img src="/extensions/vLaTeX/cache/latex_4d662ed5ce56aeded1feedc49769df6f.png" title="\bullet\stackrel{e}{\to}\bullet" style="vertical-align:-20%;" class="tex" alt="\bullet\stackrel{e}{\to}\bullet" /> <p>without hidden antiparallel identity morphisms stretched between the two copies of <img src="/extensions/vLaTeX/cache/latex_2668009033d84cdc8d16f29c11ae03fd.png" title="\bullet" style="vertical-align:-20%;" class="tex" alt="\bullet" />. If this were a category, I would probably insist that there should be hidden antiparallel identity morphisms.</p> <p>Second, when talking about commuting diagrams, we are almost always talking about functors from a free category, in which case we can mentally switch to thinking about graph morphisms due to the adjunction. This is pretty cool. Commuting diagrams in terms of graph morphisms is something I think "scientists and engineers" could really internalize easily, so this gives a nice segue to adjunctions. For the record, I think most scientists and engineers can easily internalize commuting diagrams in terms of functors too as long as they avoid the trap of thinking a diagram has to be a category (like I did!).</p> <p>A part of me still wishes we could define a diagram as follows:</p> <blockquote> <p>Given a graph <img src="/extensions/vLaTeX/cache/latex_0643966166b879e3e23697a6b846efd7.png" title="G" style="vertical-align:-20%;" class="tex" alt="G" /> and a category <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" />, a <strong>diagram</strong> in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> is a functor <img src="/extensions/vLaTeX/cache/latex_0d1483327531dd4153e5ea333e6a3753.png" title="F(G)\to C" style="vertical-align:-20%;" class="tex" alt="F(G)\to C" />.</p> <p>A <strong>commutative diagram</strong> is a diagram <img src="/extensions/vLaTeX/cache/latex_09ac822acf7c1965b64c285756e7b0a9.png" title="F(G)\to P\to C" style="vertical-align:-20%;" class="tex" alt="F(G)\to P\to C" /> which factors through a preorder <img src="/extensions/vLaTeX/cache/latex_0827b2155898e7e29e51b4330f4c0e75.png" title="P" style="vertical-align:-20%;" class="tex" alt="P" />.</p> </blockquote> <p>I like this because diagrams are almost always (I can't think of a counter example) functors from free categories for some graph. Defining it this way allows the equivalent definition stated via the adjunction:</p> <blockquote> <p>Given a graph <img src="/extensions/vLaTeX/cache/latex_0643966166b879e3e23697a6b846efd7.png" title="G" style="vertical-align:-20%;" class="tex" alt="G" /> and a category <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" />, a <strong>diagram</strong> in <img src="/extensions/vLaTeX/cache/latex_6b9aaf533fd24793040c8ec9ce302c75.png" title="U(C)" style="vertical-align:-20%;" class="tex" alt="U(C)" /> is a graph morphism <img src="/extensions/vLaTeX/cache/latex_27cc390aec1b1c61f926a3fbe835813e.png" title="G\to U(C)" style="vertical-align:-20%;" class="tex" alt="G\to U(C)" />.</p> <p>A <strong>commutative diagram</strong> is a diagram <img src="/extensions/vLaTeX/cache/latex_09ac822acf7c1965b64c285756e7b0a9.png" title="F(G)\to P\to C" style="vertical-align:-20%;" class="tex" alt="F(G)\to P\to C" /> which factors through a preorder <img src="/extensions/vLaTeX/cache/latex_0827b2155898e7e29e51b4330f4c0e75.png" title="P" style="vertical-align:-20%;" class="tex" alt="P" />.</p> </blockquote> <p>In fact, that last sentence could probably be improved, because I'm willing to bet that any preorder is a quiver for some graph, so it could probably be recast as a special kind of graph morphism.</p> <p>There is one more worry on my mind though...</p> <p>Back <a href="http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=332&Focus=7112#Comment_7112">here</a>, I gave two definitions:</p> <hr /> <p><strong>Definition #1:</strong></p> <p>Given categories <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> and <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" />, a <strong>functor</strong> <img src="/extensions/vLaTeX/cache/latex_4ab4a8e1b1343def201689148996e8cd.png" title="F:C\to D" style="vertical-align:-20%;" class="tex" alt="F:C\to D" /> is a map that sends each object <img src="/extensions/vLaTeX/cache/latex_0e29540f708ca8b32bde8e2c33e97582.png" title="x" style="vertical-align:-20%;" class="tex" alt="x" /> in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> to an object <img src="/extensions/vLaTeX/cache/latex_91281b8b55384164c30dff8ad67a7219.png" title="F(x)" style="vertical-align:-20%;" class="tex" alt="F(x)" /> in <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" /> and each morphism <img src="/extensions/vLaTeX/cache/latex_fda7a63267491b34b45b8d7b684bb1f8.png" title="f:x\to y" style="vertical-align:-20%;" class="tex" alt="f:x\to y" /> in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> to a morphism <img src="/extensions/vLaTeX/cache/latex_ddd586710e765e3df32ec4ac5dfb8f2c.png" title="F(f):F(x)\to F(y)" style="vertical-align:-20%;" class="tex" alt="F(f):F(x)\to F(y)" /> in <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" /> such that:</p> <ol> <li><img src="/extensions/vLaTeX/cache/latex_5ec3dcd4fa96727f6008946ebbeebfa2.png" title="F" style="vertical-align:-20%;" class="tex" alt="F" /> preserves compositions,</li> <li><img src="/extensions/vLaTeX/cache/latex_5ec3dcd4fa96727f6008946ebbeebfa2.png" title="F" style="vertical-align:-20%;" class="tex" alt="F" /> preserves identities. </ol> <p><strong>Definition #2:</strong></p> <p>Given categories <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> and <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" />, a <strong>functor</strong> <img src="/extensions/vLaTeX/cache/latex_4ab4a8e1b1343def201689148996e8cd.png" title="F:C\to D" style="vertical-align:-20%;" class="tex" alt="F:C\to D" /> is a map that sends each object <img src="/extensions/vLaTeX/cache/latex_0e29540f708ca8b32bde8e2c33e97582.png" title="x" style="vertical-align:-20%;" class="tex" alt="x" /> in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> to an object <img src="/extensions/vLaTeX/cache/latex_91281b8b55384164c30dff8ad67a7219.png" title="F(x)" style="vertical-align:-20%;" class="tex" alt="F(x)" /> in <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" /> and each morphism <img src="/extensions/vLaTeX/cache/latex_fda7a63267491b34b45b8d7b684bb1f8.png" title="f:x\to y" style="vertical-align:-20%;" class="tex" alt="f:x\to y" /> in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> to a morphism <img src="/extensions/vLaTeX/cache/latex_ddd586710e765e3df32ec4ac5dfb8f2c.png" title="F(f):F(x)\to F(y)" style="vertical-align:-20%;" class="tex" alt="F(f):F(x)\to F(y)" /> in <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" /> such that each commuting diagram in <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align:-20%;" class="tex" alt="C" /> maps to a commuting diagram in <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align:-20%;" class="tex" alt="D" />.</p> <hr /> <p>Mike responded <a href="http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=332&Focus=7140#Comment_7140">here</a>:</p> <blockquote> <p>@Eric: I would consider your definition #2 to be unbiased, and to include definition #1. The two axioms in definition #1 are both particular sorts of commuting diagrams: a composition of arrows is a commuting triangle, and an identity is a commuting loop.</p> </blockquote> <p>I interpreted Mike's response to mean Definition #1 and Definition #2 were the same, but all Mike really said was #2 includes #1. For them to be the same, #1 would also have to include #2. I'm afraid that might not be the case. I'm afraid #2 might be stronger than #1. I'm not sure, but I think with #2, we WOULD have the image always being a subcategory. What do you think?</p> ]]> Todd_Trimble comments on "functor" (7464) https://nforum.ncatlab.org/discussion/332/?Focus=7464#Comment_7464 2010-04-03T00:03:18+00:00 2023-09-24T23:56:33+00:00 Todd_Trimble https://nforum.ncatlab.org/account/24/ That sounds good to me, Mike. Eric, my own take is that different terms for (roughly) the same concept can convey slightly different senses or "moods" or emphases, just as in ordinary ...

That sounds good to me, Mike.

Eric, my own take is that different terms for (roughly) the same concept can convey slightly different senses or "moods" or emphases, just as in ordinary language use. For example, I would sooner say "a diagram in $D$ is a functor $F: J \to D$, where $J$ is small. A representation of $C$ is a functor whose domain is $C$." There's a difference in sense.

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Mike Shulman comments on "functor" (7461) https://nforum.ncatlab.org/discussion/332/?Focus=7461#Comment_7461 2010-04-02T22:01:27+00:00 2023-09-24T23:56:33+00:00 Mike Shulman https://nforum.ncatlab.org/account/3/ What about "A diagram is a functor. When we ask whether a diagram commutes, we are usually thinking only of diagrams whose source is the free category on a graph, in which case we say the ...

What about "A diagram is a functor. When we ask whether a diagram commutes, we are usually thinking only of diagrams whose source is the free category on a graph, in which case we say the diagram commutes iff it factors through the free preorder on that graph (or, equivalently, through any preorder). Note that by adjunction, a functor $Free(G) \to C$ is the same as a graph morphism $G\to Underlying(C)$, i.e. a 'labeling' of the vertices and edges of G by objects and morphisms in C. This labeling is what we usually draw on the page when we talk about commutative diagrams."

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Eric comments on "functor" (7456) https://nforum.ncatlab.org/discussion/332/?Focus=7456#Comment_7456 2010-04-02T20:30:47+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ Hmm... I'm not a big fan of definitions like: A diagram is simply a functor . Period. A representation is simply a functor . Period. That makes me wonder, "Is a diagram simply a ...

Hmm... I'm not a big fan of definitions like:

• A diagram is simply a functor $F:C\to D$. Period.
• A representation is simply a functor $F:C\to D$. Period.

That makes me wonder, "Is a diagram simply a representation? Why have three words for the same thing?"

In this case, I probably would prefer to restrict things just a little unless there is a good reason not to. That way I can safely keep in the back of my head "a diagram is basically a graph morphism". This is what it means for all practical purposes.

Should we take a vote on it? :)

Thanks again everyone for your help. I just redrew my favorite example, but now starting with a graph $G$ given by $X\stackrel{f}{\to} Y$

and a graph $L$ given by $\stackrel{e}{\circlearrowleft}$.

From here we can construct the free category of the loop $F(L)$ and project this back down to a graph $U\circ F(L)$ with one node and an infinite number of directed edges, one for each morphism $1$, $e$, $e^2$, etc.

A diagram is a graph morphism $G\to U\circ F(L)$. Of course $f\mapsto e^n$ for some $n\ge 0$ and there are no hidden identities etc forcing $f\mapsto 1$.

Mystery solved :)

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Todd_Trimble comments on "functor" (7451) https://nforum.ncatlab.org/discussion/332/?Focus=7451#Comment_7451 2010-04-02T20:09:31+00:00 2023-09-24T23:56:33+00:00 Todd_Trimble https://nforum.ncatlab.org/account/24/ Yes, that formulation doesn't mention that adjunction. Graph morphisms would be special cases. But those are the cases which generally arise in practice when we wave our hands and point to or draw ...

Yes, that formulation doesn't mention that adjunction. Graph morphisms would be special cases. But those are the cases which generally arise in practice when we wave our hands and point to or draw commutative diagrams, as one can verify by opening just about any book. :-)

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Eric comments on "functor" (7448) https://nforum.ncatlab.org/discussion/332/?Focus=7448#Comment_7448 2010-04-02T20:00:06+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ @Todd: Ok, but if is not free, do we need to abandon the adjunction? I would like to be able to think in terms of graph morphisms, but in the more general case, would graph morphisms just be special ...

@Todd: Ok, but if $J$ is not free, do we need to abandon the adjunction? I would like to be able to think in terms of graph morphisms, but in the more general case, would graph morphisms just be special cases?

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Todd_Trimble comments on "functor" (7445) https://nforum.ncatlab.org/discussion/332/?Focus=7445#Comment_7445 2010-04-02T19:52:16+00:00 2023-09-24T23:56:33+00:00 Todd_Trimble https://nforum.ncatlab.org/account/24/ Eric: you could rephrase that way, but now that this discussion is at last converging toward consensus (yay!), I think the issue of whether to use free categories or not becomes a bit subsidiary and ...

Eric: you could rephrase that way, but now that this discussion is at last converging toward consensus (yay!), I think the issue of whether to use free categories or not becomes a bit subsidiary and can be relegated to a comment.

In other words, you could define a diagram as we currently have it in the Lab, as a functor $F: J \to C$, and say the diagram commutes if it factors through a preorder $P$. (This $P$ can be taken to be the preorder reflection of $J$ -- offhand I'm not sure where that would link to.) This uses a few less words than if you brought in free categories or whatever, and is somewhat more general. But I still think Mike's earlier words on this still have a certain legitimacy and deserve to be noted as a comment in the article.

I lean slightly toward the more general formulation, but either definition is okay with me; the definition you want to use Eric is certainly defensible. At this point I can bow out of the conversation. :-)

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Eric comments on "functor" (7444) https://nforum.ncatlab.org/discussion/332/?Focus=7444#Comment_7444 2010-04-02T19:48:04+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ @Harry: No worries about that. The help you give more than makes up for things like that. Plus, I somewhat disagree with Todd. Being unfriendly towards engineers actually puts you in very good ...

@Harry: No worries about that. The help you give more than makes up for things like that. Plus, I somewhat disagree with Todd. Being unfriendly towards engineers actually puts you in very good company in the mathematics community (present company excluded!) :)

As I mentioned before, since you are not a "scientist nor engineer", you do not have a say in what they find easier :) The most you can say is that it is easier for you and that is great. I've been around long enough and I've known enough "scientists and engineers" to say with some confidence that most would be much more comfortable working with graphs, since graphs pervade almost all scientific disciplines.

Besides, in my suggested modification to the nLab above, the functor aspect was front and center and was part of the definition. The relation to graph morphisms was secondary.

Or are you hesitant about using the word "graph" in the definition? If that is the case, then I'd disagree. I think it is perfectly acceptable for the word graph to appear in the definition of a diagram.

What I'd really like to know is whether my suggested modification is acceptable to the experts. Is it?

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Harry Gindi comments on "functor" (7443) https://nforum.ncatlab.org/discussion/332/?Focus=7443#Comment_7443 2010-04-02T19:33:07+00:00 2023-09-24T23:56:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ @Eric: I apologize if that came off a bit rude. It was not my intention. The main problem that I see with using the "functors are maps of graphs" approach is that it is a lot more ...

@Eric: I apologize if that came off a bit rude. It was not my intention.

The main problem that I see with using the "functors are maps of graphs" approach is that it is a lot more confusing and complicated. The real benefit from the diagram approach comes from things like limits and colimits, where the choice of index category only matters for the shape. However, in practice, it still feels a lot easier to just be careful with your definition of limit/colimit, which you can apply to arbitrary functors (provided they exist).

I prefer to reserve statements about commutativity to actual diagrams that I can draw (or at least pieces of diagrams, if they're infinite). All of these statements about posets etc are interesting, but we all have an intuitive notion of what it means for a diagram to commute. I submit to you that the notion of an abstract diagram as a map on graphs, let alone any definition of what it means for such a creature to commute, is certainly more difficult to work with than saying that a functor is a map on the set of objects and the set of arrows preserving identities and composition. Even if we want to use an unbiased definition, it's still the same. That is, we just say that F preserves n-ary composites for all n in N (notably including 0, where the 0ary composite at an object is the identity).

(By the way, defining the identity to be the 0-ary composite doesn't seem right, since the 0-ary composite does not have an explicit domain/codomain. This means that we have to define a notion of a 0-ary composite at an object, which is much less parsimonious than just splitting off the case of n=0.)

Edit: Eric, I am against changing the definition of a diagram. The whole beauty of the standard approach is not that we can describe categories as graphs, but that we can describe a diagram as a functor!

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Eric comments on "functor" (7439) https://nforum.ncatlab.org/discussion/332/?Focus=7439#Comment_7439 2010-04-02T19:07:37+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ The nLab says: A diagram in a category is simply a functor . The category is called the shape or index category of the diagram, and is typically understood to be a small category. Does ...

The nLab says:

A diagram in a category $C$ is simply a functor $F:J\to C$.

The category $J$ is called the shape or index category of the diagram, and is typically understood to be a small category.

Does $J$ need to be a free category?

If so, I think it would be helpful to explicitly state it like I did above:

Given a graph $G$ and a category $C$, a diagram in $C$ is a functor $F(G)\to C$.

A commutative diagram is a diagram $F(G)\to P\to C$ which factors through a preorder $P$.

This would be much clearer to me. What do you think? Can we change the nLab?

This would also lead directly to the statement:

A diagram may also be thought of (via adjunction) as a graph morphism $G\to U(C)$.

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Todd_Trimble comments on "functor" (7434) https://nforum.ncatlab.org/discussion/332/?Focus=7434#Comment_7434 2010-04-02T17:46:28+00:00 2023-09-24T23:56:33+00:00 Todd_Trimble https://nforum.ncatlab.org/account/24/ This comment is invalid XHTML+MathML+SVG; displaying source. &lt;div&gt; &lt;p&gt;I think my last comment addressed some of your questions, except for where you tender this ... This comment is invalid XHTML+MathML+SVG; displaying source. <div> <p>I think my last comment addressed some of your questions, except for where you tender this suggestion:</p> <blockquote> A commutative diagram is a diagram in which F(G) is a poset. </blockquote> <p>I'm glad you brought that up. I wouldn't agree with it though, because it rules out the possibility of having any loops in <img src="/extensions/vLaTeX/cache/latex_0643966166b879e3e23697a6b846efd7.png" title="G" style="vertical-align: -20%;" class="tex" alt="G"/>.</p> <p>It <em>would</em> be correct to say, however, that a diagram <img src="/extensions/vLaTeX/cache/latex_57692eca3f93f29b3582d21504dcd5f8.png" title="Free(G) \to D" style="vertical-align: -20%;" class="tex" alt="Free(G) \to D"/> commutes if it factors through a poset (or preorder): can be expressed as a composite functor <img src="/extensions/vLaTeX/cache/latex_c7e7213bca879361643e07fc87a2ae89.png" title="Free(G) \to P \to D" style="vertical-align: -20%;" class="tex" alt="Free(G) \to P \to D"/> where <img src="/extensions/vLaTeX/cache/latex_0827b2155898e7e29e51b4330f4c0e75.png" title="P" style="vertical-align: -20%;" class="tex" alt="P"/> is a preorder. That's a sort of slick and tricky way to restate the condition I mentioned earlier: that whenever <img src="/extensions/vLaTeX/cache/latex_43c09e88e076b2d340501a0744aa1639.png" title="f" style="vertical-align: -20%;" class="tex" alt="f"/> and <img src="/extensions/vLaTeX/cache/latex_333d2017e7f892329a2362f4087a4bd3.png" title="g" style="vertical-align: -20%;" class="tex" alt="g"/> are parallel morphisms of <img src="/extensions/vLaTeX/cache/latex_bfcfae52c824f60bfd5e0a2b0950e6fa.png" title="Free(G)" style="vertical-align: -20%;" class="tex" alt="Free(G)"/> (i.e., paths in <img src="/extensions/vLaTeX/cache/latex_0643966166b879e3e23697a6b846efd7.png" title="G" style="vertical-align: -20%;" class="tex" alt="G"/> which start at the same place and end at the same place), that <img src="/extensions/vLaTeX/cache/latex_85fcb2cf89e0d4077de91402a06775cd.png" title="F(f) = F(g)" style="vertical-align: -20%;" class="tex" alt="F(f) = F(g)"/>. There are other ways of putting it too.</p> <p>Edit: Eric #49 -- that's right.</p> </div>]]> Eric comments on "functor" (7433) https://nforum.ncatlab.org/discussion/332/?Focus=7433#Comment_7433 2010-04-02T17:35:48+00:00 2023-09-24T23:56:33+00:00 Eric https://nforum.ncatlab.org/account/5/ This comment is invalid XML; displaying source. &lt;p&gt;Thanks again Todd! Quick note...&lt;/p&gt; &lt;p&gt;When I said &quot;same&quot;, I didn&#039;t mean ... This comment is invalid XML; displaying source. <p>Thanks again Todd! Quick note...</p> <p>When I said "same", I didn't mean "literally the same", I should have said something like "Was it an adjunction that <a href="http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=332&Focus=7340#Comment_7340">Mike was talking about here</a>? Was he simply expressing the same concept in different contexts implicitly using an adjunction? Or were the differences more than that?"</p> <p>I think you just answered in the affirmative. Mike was implicitly using the adjunction (although he explicitly mentioned it in his next comment). Right?</p> ]]> Todd_Trimble comments on "functor" (7432) https://nforum.ncatlab.org/discussion/332/?Focus=7432#Comment_7432 2010-04-02T17:30:15+00:00 2023-09-24T23:56:33+00:00 Todd_Trimble https://nforum.ncatlab.org/account/24/ This comment is invalid XHTML+MathML+SVG; displaying source. &lt;div&gt; &lt;blockquote&gt; So our domain had two objects and three morphisms unless we are thinking of it as just a ... This comment is invalid XHTML+MathML+SVG; displaying source. <div> <blockquote> So our domain had two objects and three morphisms unless we are thinking of it as just a graph. </blockquote> <p>That's right.</p> <blockquote> How many objects does the codomain have? How many morphisms does the codomain have? </blockquote> <p>I want to say it doesn't matter -- it depends. In my original example, the codomain was <img src="/extensions/vLaTeX/cache/latex_4981dace9a12cac85b96282cad1141cb.png" title="Set" style="vertical-align: -20%;" class="tex" alt="Set"/>, so there are jillions of each. But from that large container we picked just the bit that we wanted: in the running example, we assigned $\mathbb{N}$ to both of the objects of the domain, and assigned the successor function to the non-identity arrow. We could have chosen different assignments.</p> <p>It may help to think of it this way: the domain <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align: -20%;" class="tex" alt="C"/> gives you the underlying shape of the diagram, and the diagram itself is that shape together with a labeling of the parts of that shape by data which belongs to the codomain; the functor assignments give precisely that labeling.</p> <blockquote> The identity morphisms have to go somewhere (unless it was just a graph all along). </blockquote> <p>That's right: they are perforce sent to (or assigned) identity morphisms.</p> <p>I think I see where you might think this heads you toward trouble, but this is where we need very clear language to keep all the components of the picture straight and disambiguated. Assuming we are for the moment still thinking of diagrams as functors, the definition of what it means for a diagram to commute is very carefully stated: whenever <img src="/extensions/vLaTeX/cache/latex_43c09e88e076b2d340501a0744aa1639.png" title="f" style="vertical-align: -20%;" class="tex" alt="f"/> and <img src="/extensions/vLaTeX/cache/latex_333d2017e7f892329a2362f4087a4bd3.png" title="g" style="vertical-align: -20%;" class="tex" alt="g"/> are two parallel morphisms belonging to the <em>domain</em> ("parallel" meaning: have the same domain and have the same codomain), the diagram (functor) <img src="/extensions/vLaTeX/cache/latex_5ec3dcd4fa96727f6008946ebbeebfa2.png" title="F" style="vertical-align: -20%;" class="tex" alt="F"/> sends them to the same map: <img src="/extensions/vLaTeX/cache/latex_85fcb2cf89e0d4077de91402a06775cd.png" title="F(f) = F(g)" style="vertical-align: -20%;" class="tex" alt="F(f) = F(g)"/>. The diagram-picture comes in two components, really: there is the underlying scheme or shape -- that's the domain -- and then on top of that is superimposed, so to speak, a labeling of the parts of the domain by codomain data; the functor <img src="/extensions/vLaTeX/cache/latex_5ec3dcd4fa96727f6008946ebbeebfa2.png" title="F" style="vertical-align: -20%;" class="tex" alt="F"/> itself performs the labeling.</p> <p>Now: check the example against this carefully-worded definition. You'll see that the diagram commutes, no problem. As long as you keep the two aforementioned two components of the picture straight, you won't fall into paradox.</p> <blockquote> Also, I am confused because a diagram can apparently be a functor or it can be a morphism of graphs. Through adjunction, these two are related, but are they the same? </blockquote> <p>Not literally the same, no: a graph morphism <img src="/extensions/vLaTeX/cache/latex_375ac6db8532ffc69d178257a0a4cfbb.png" title="G \to U(D)" style="vertical-align: -20%;" class="tex" alt="G \to U(D)"/> into the underlying graph of a category <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align: -20%;" class="tex" alt="D"/> is not literally, or in terms of strict logic, the same as a functor <img src="/extensions/vLaTeX/cache/latex_cadb5f19893654cd2fa1ccb665e136a0.png" title="F(G) \to D" style="vertical-align: -20%;" class="tex" alt="F(G) \to D"/> from the free category on a graph <img src="/extensions/vLaTeX/cache/latex_0643966166b879e3e23697a6b846efd7.png" title="G" style="vertical-align: -20%;" class="tex" alt="G"/>. But you can go back and forth freely, and these notions carry essentially the same information.</p> <blockquote> When you guys mention there are two ways to think about diagrams, are these two ways the same expressed differently, or are they truly different? When Mike first mentioned the two meanings, I thought they were different. </blockquote> <p>That's right -- they are different. But spiritually speaking, not different in any very significant sense. To be precise, the meaning of diagram I was using, where we use any category <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align: -20%;" class="tex" alt="C"/> as domain and <img src="/extensions/vLaTeX/cache/latex_ca43fb5496104dcafda44acbe4014b0e.png" title="C" style="vertical-align: -20%;" class="tex" alt="C"/>-diagrams in <img src="/extensions/vLaTeX/cache/latex_2741ee6cf4a5aa48b2ee29b3b0fee62e.png" title="D" style="vertical-align: -20%;" class="tex" alt="D"/> are functors <img src="/extensions/vLaTeX/cache/latex_b8084f37bc2b84e2ef0a212ed3bc8bd3.png" title="C \to D" style="vertical-align: -20%;" class="tex" alt="C \to D"/>, is strictly more general than Mike's meaning (where we use graph morphisms <img src="/extensions/vLaTeX/cache/latex_375ac6db8532ffc69d178257a0a4cfbb.png" title="G \to U(D)" style="vertical-align: -20%;" class="tex" alt="G \to U(D)"/>, or equivalently functors <img src="/extensions/vLaTeX/cache/latex_cadb5f19893654cd2fa1ccb665e136a0.png" title="F(G) \to D" style="vertical-align: -20%;" class="tex" alt="F(G) \to D"/> coming out of free categories). But Mike was arguing that in actual practice, we just about always use the meaning he gave, and the one I gave has extra somewhat spurious generality not needed in practice. I believe I agree with that, and neither of us seem to be too exercised about the distinction.</p> <p>@Harry: please don't speak down to Eric. He is trying, and I believe I am correct in saying that while he knows a lot of mathematics, he is an engineer by training, and is coming from a different scientific culture. I think we all agree that mathematics is hard, and we need to be patient toward those who are struggling and have the courage to expose that fact. (You may not be a professional teacher now, but you are probably contemplating such a track, and it's never too soon to train oneself to become a helping teacher.)</p> </div>]]>