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• CommentRowNumber1.
• CommentAuthorzskoda
• CommentTimeSep 2nd 2010
• (edited Sep 2nd 2010)

Using codecogs recipe and ascii table I wrote short entries fork and split equalizer. For those who distinguish fork and cofork, I have hard time remembering which one is which one.

By the way, nForum is today having lots of problems on my computer, it asks for human recognition, it bails out my automatically remembered password many times out and resets the settings for markdown when writing etc. often.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeSep 3rd 2010

You should report these problems here (and read the top post).

• CommentRowNumber3.
• CommentAuthorEric
• CommentTimeSep 3rd 2010
• (edited Sep 3rd 2010)

Note: Whoa. Markdown doesn’t seem to be working (in preview only, apparently). I have checked “Markdown+Itex”.

I was going to add a note at fork that it was a cone over the diagram consisting of 2 parallel morphisms. But reviewing cone, it says that the original diagram must already be commuting (unless I’m misreading it).

In category theory a cone over a commuting diagram is an object equipped with morphisms from it into each vertex of the diagram, such that all new diagrams arising this way commute.

The parallel morphisms do not already need to be commuting, right? Could this be rewritten as

In category theory a cone over a diagram is an object equipped with morphisms from it into each vertex of the original diagram such that this new diagram commutes.

If so, then a fork is a cone over the parallel morphisms and a cofork would be a cocone. Adding a note like this might be good (if correct!).

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeSep 3rd 2010
• (edited Sep 3rd 2010)

A cone over a commuting diagram??? WTF?

“Commuting” there should be stricken. Then, to say that the “new diagram” should commute is confusing (actually, wrong), but the idea is that the shape of the new diagram (as a category) is obtained by adjoining an initial object, and the new diagram maps the initial object to the cone vertex. Not sure of the best way of putting that “informally”.

• CommentRowNumber5.
• CommentAuthorEric
• CommentTimeSep 3rd 2010
• (edited Sep 3rd 2010)

the idea is that the shape of the new diagram (as a category)

But diagrams need not be categories, right?

Then, to say that the “new diagram” should commute is confusing (actually, wrong)

There’s got to be some nice “arrow theoretic” way to say this. Could you say something along the lines that a cone $C$ over a diagram $D$ is a commuting diagram such that $D$ is “somehow” a subdiagram? Or something…

Just thinking out loud…

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeSep 3rd 2010

But diagrams need not be categories, right?

no, as they are functors, but the shape of a diagram is* a category ($D$, say). Add an extra object 0, and a unique arrow from this object to any other object. If you follow the axioms, then any path in this modified category $D[0]$ from 0 to any other is equal to any other such path. Note that this doesn’t force any other arrows (or paths, if you like) to be equal. Then a cone is a way of extending the functor $D \to C$ to $D[0]$ - if any exist at all.

*)well, it could be a graph, but then secretly we take the free category on that graph.

• CommentRowNumber7.
• CommentAuthorEric
• CommentTimeSep 3rd 2010
• (edited Sep 3rd 2010)

no, as they are functors, but the shape of a diagram is* a category (D, say).

Sorry. Sure. A diagram is a functor $F:J\to D$. The shape $J$ is a category, but the image of $J$ in $D$ need not be a subcategory of $D$. That is what I meant. I misread Todd’s comment (even though it was stated perfectly clearly). Sorry about that.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeSep 3rd 2010

• CommentRowNumber9.
• CommentAuthorzskoda
• CommentTimeSep 3rd 2010
• (edited Sep 3rd 2010)

Well, you added a parallel pair. Fork informally consists of a parallel pair together with a cone. Edit: Oh no, I misunderstood which part you added… :)

• CommentRowNumber10.
• CommentAuthorTobyBartels
• CommentTimeSep 3rd 2010
• (edited Sep 3rd 2010)

@ Todd #4

I’ve fixed this at cone. I believe that I got across the idea of what should commute by stressing the word ‘new’.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeSep 3rd 2010

Toby: right. It’s correct to use the phrase “new triangle” rather than “new diagram”.

• CommentRowNumber12.
• CommentAuthorEric
• CommentTimeSep 6th 2010

Toby: right. It’s correct to use the phrase “new triangle” rather than “new diagram”.

I’m still confused by this. It is probably just a matter of semantics. The result of adding a new object and morphisms from that object to all original objects, to me, amounts to constructing a new diagram. This new diagram commutes. Where is the flaw? I’m not challenging the idea, but would rather like to understand it even if only to get the semantics right.

• CommentRowNumber13.
• CommentAuthorDavidRoberts
• CommentTimeSep 6th 2010

Actually it is a bit ambiguous to say ’a diagram’ commutes. The ’commuting’ refers to composites of arrows being equal, but you have to specify which composites of which arrows. And cones only require that certain triangles commute (even in the case of a cone over a parallel pair $f,g:X\to Y$: there are two arrows from the vertex $x:V \to X$ and $y:V \to Y$. That ’certain triangles’ commute means that $y = f\circ x$ and $y = g\circ x$…)

• CommentRowNumber14.
• CommentAuthorTodd_Trimble
• CommentTimeSep 6th 2010
• (edited Sep 6th 2010)

amounts to constructing a new diagram

Correct.

This new diagram commutes.

No. Consider the example where the original diagram is given by a parallel pair of distinct morphisms $g, h: b \stackrel{\to}{\to} c$. A cone consists of an object $a$ and morphism $f: a \to b$ such that $g f = h f$. But the new diagram still does not commute.

Why? Look at the definition in commutative diagram. It says that for any two vertices in the diagram $x$, $y$, any two paths through the diagram from $x$ to $y$ must yield the same morphism. But for $x = b$, $y = c$, this does not hold if we take one path to be $f$ and the other $g$.

In other words, if the old diagram doesn’t commute, then the new diagram won’t either, for the exact same reason. But, any two paths which start from the newly adjoined vertex and end at the same place do commute. So it’s a restricted commutativity. And it suffices to consider triangles, one of whose vertices is the new vertex.

Does it make sense now? It’s not just semantics.

• CommentRowNumber15.
• CommentAuthorEric
• CommentTimeSep 6th 2010

Yes. Thanks for explaining. Sorry for being dense.