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1. • CommentRowNumber1.
   • CommentAuthorEric
   • CommentTimeJul 25th 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexA [[diagram]] is a functor $D:J\to C$. As I've learned here on the nForum, although $J$ is a category, the image $F(J)$ need not be a subcategory of $C$. A good example is where $J$ is given by $a\stackrel{f}{\to} b$ and we set $F(a) = F(b)$. The resulting loop diagram is not a subcategory of $C$ because the diagram doesn't contain powers of $f$, i.e. $f^2$, $f^3$, etc. So, I suppose this means when we are talking about limits of finite diagrams, we can't just dream up any shape in $C$. Our shape has to come from the image of some category $J$, right? In particular, does this mean that any diagram we want to take a limit of must contain all composites? For example, is $a \stackrel{f}{\to} b\stackrel{g}{\to} c$ a valid diagram we can take a limit of or would we need to include $a\stackrel{f\circ g}{\to} c$ explicitly in the diagram? I've been practicing with finite limits and thought I'd try the loop diagram $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$ but paused because I couldn't think of a category $J$ for which the above diagram would be the image so I wasn't sure if this is even a valid diagram to try to take a limit of. Hence, I was wondering if there are special allowed shapes for diagrams or if they can be pretty much arbitrary. In particular, is the loop $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$ a diagram we can take a limit of? I'll go ahead and try it just to see what happens.

   A diagram is a functor $D:J\to C$.

   As I’ve learned here on the nForum, although $J$ is a category, the image $F(J)$ need not be a subcategory of $C$.

   A good example is where $J$ is given by $a\stackrel{f}{\to} b$ and we set $F(a) = F(b)$. The resulting loop diagram is not a subcategory of $C$ because the diagram doesn’t contain powers of $f$, i.e. $f^2$, $f^3$, etc.

   So, I suppose this means when we are talking about limits of finite diagrams, we can’t just dream up any shape in $C$. Our shape has to come from the image of some category $J$, right?

   In particular, does this mean that any diagram we want to take a limit of must contain all composites?

   For example, is

   $a \stackrel{f}{\to} b\stackrel{g}{\to} c$

   a valid diagram we can take a limit of or would we need to include $a\stackrel{f\circ g}{\to} c$ explicitly in the diagram?

   I’ve been practicing with finite limits and thought I’d try the loop diagram

   $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$

   but paused because I couldn’t think of a category $J$ for which the above diagram would be the image so I wasn’t sure if this is even a valid diagram to try to take a limit of.

   Hence, I was wondering if there are special allowed shapes for diagrams or if they can be pretty much arbitrary.

   In particular, is the loop

   $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$

   a diagram we can take a limit of?

   I’ll go ahead and try it just to see what happens.

2. • CommentRowNumber2.
   • CommentAuthorTobyBartels
   • CommentTimeJul 25th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>In particular, does this mean that any diagram we want to take a limit of must contain all composites? No, not at all! You can just go ahead and work out the [[limit]] of it (if one exists) anyway. In other words, it is *not* correct to think that we take limits only of subcategories, so the fact that not every diagram defines a subcategory is irrelevant. Of course, we *can* take limits of subcategories; if $J$ is a subcategory of $C$, then the inclusion functor $D\colon J \to C$ is a diagram in $C$, so we can consider the limit of that. But that is only a *special case*. When you just draw a finite diagram, you're usually not drawing a subcategory. What's worse, you're usually not drawing the image of a functor either! Instead, you're usually drawing the image of a map from a [[quiver]]. In the case of $$a \stackrel{f}{\to} b\stackrel{g}{\to} c$$ that is exactly what is going on. Ignoring the labels ($a$, $b$, $c$, $f$, and $g$), you've drawn a quiver with $3$ vertices and $2$ edges, but this generates a category with $3$ objects and $6$ morphisms. So if you think of the diagram as a functor, then yes, $f ; g$ is part of its image, as are $\id_a$, $\id_b$, and $\id_c$. However, you are not obligated to draw them in. And the great thing about diagrams from quivers is that you can take the same concrete definition of [[limit]] and apply it directly to them, without bothering about the missing composites and identity morphisms; they won't change the result! This is a theorem, not something that's supposed to be immediately obvious. Another theorem is that a diagram from a finite quiver has a limit in any finitely complete category, even if the free category on that quiver is *not* finite. We see this with $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$ where you have a quiver with $3$ vertices and $3$ edges, which generates a category with $3$ objects and $3 \omega$ morphisms. In $Fin Set$, its limit is $$ \{ x\colon a, y\colon b, z\colon c \;|\; y = f(x), z = g(y), x = h(z) \} ,$$ which hopefully looks obvious once you see it, and basically shows how you'd construct the limit out of finite products and equalisers in any finitely complete category. There are infinitely many more conditions that you *could* write on the right side of that bar, but they all follow from the $3$ above.

   In particular, does this mean that any diagram we want to take a limit of must contain all composites?

   No, not at all! You can just go ahead and work out the limit of it (if one exists) anyway.

   In other words, it is not correct to think that we take limits only of subcategories, so the fact that not every diagram defines a subcategory is irrelevant.

   Of course, we can take limits of subcategories; if $J$ is a subcategory of $C$, then the inclusion functor $D\colon J \to C$ is a diagram in $C$, so we can consider the limit of that. But that is only a special case.

   When you just draw a finite diagram, you’re usually not drawing a subcategory. What’s worse, you’re usually not drawing the image of a functor either! Instead, you’re usually drawing the image of a map from a quiver.

   In the case of

   $$a \stackrel{f}{\to} b\stackrel{g}{\to} c$$

   that is exactly what is going on. Ignoring the labels ($a$, $b$, $c$, $f$, and $g$), you’ve drawn a quiver with $3$ vertices and $2$ edges, but this generates a category with $3$ objects and $6$ morphisms. So if you think of the diagram as a functor, then yes, $f ; g$ is part of its image, as are $\id_a$, $\id_b$, and $\id_c$. However, you are not obligated to draw them in.

   And the great thing about diagrams from quivers is that you can take the same concrete definition of limit and apply it directly to them, without bothering about the missing composites and identity morphisms; they won’t change the result! This is a theorem, not something that’s supposed to be immediately obvious. Another theorem is that a diagram from a finite quiver has a limit in any finitely complete category, even if the free category on that quiver is not finite.

   We see this with

   $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$

   where you have a quiver with $3$ vertices and $3$ edges, which generates a category with $3$ objects and $3 \omega$ morphisms. In $Fin Set$, its limit is

   $$\{ x\colon a, y\colon b, z\colon c \;|\; y = f(x), z = g(y), x = h(z) \} ,$$

   which hopefully looks obvious once you see it, and basically shows how you’d construct the limit out of finite products and equalisers in any finitely complete category. There are infinitely many more conditions that you could write on the right side of that bar, but they all follow from the $3$ above.

3. • CommentRowNumber3.
   • CommentAuthorEric
   • CommentTimeJul 26th 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexCool. Thanks Toby :) Just to clarify, I was thinking of a diagram as a functor rather than a graph morphism although I understand there is the adjointness relation. So when I asked about composites, I should have been more clear. What I meant is that since $J$ is a category, it obviously contains all composites (and identities). Since a diagram is a functor, the image $F(J)$ should contain all composites that were already in $J$. However, I do understand that generally $F(J)$ may not contain all composites in $C$ so it may not be a subcategory of $C$. In the example I gave where $J$ is given by $a\stackrel{f}{\to} b$, the image $F(J)$ contains all composites that were already in $J$, but it does not contain all composites in $C$ when we set $F(a) = F(b)$ so that $F(f)$ is a loop, because $F(f)\circ F(f)$ is not in the diagram in $C$ so the diagram is not a subcategory. > Instead, you're usually drawing the image of a map from a [[quiver]]. >In the case of $$a \stackrel{f}{\to} b\stackrel{g}{\to} c$$ >that is exactly what is going on. Yeah, I understand this. My question was pretty clumsy, but I was thinking in terms of functors. The above is not the image of some $J$ for any functor because it doesn't contain the composite $g\circ f:a\to c$. So in a way, it is "not allowed". But as a quiver, I understand it is perfectly legit and the free category generated from this would include the composite. >And the great thing about diagrams from quivers is that you can take the same concrete definition of [[limit]] and apply it directly to them, without bothering about the missing composites and identity morphisms; they won't change the result! Very neat :) So yeah, taking limits of arbitrary quivers will provide all kinds of exercises :) Is this because of the adjoint relation? When $F:C\to D$ and $G:D\to C$ are adjoint, do $F$ and $G$ preserve all finite limits? Oh, I see the theorem you are probably referring to :) [[adjoint functor theorem]] So the right adjoint preserves limits (which I guess the free functor would be right adjoint) and the left adjoint (which would be the forgetful functor) preserves colimits. Thanks :)

   Cool. Thanks Toby :)

   Just to clarify, I was thinking of a diagram as a functor rather than a graph morphism although I understand there is the adjointness relation.

   So when I asked about composites, I should have been more clear.

   What I meant is that since $J$ is a category, it obviously contains all composites (and identities). Since a diagram is a functor, the image $F(J)$ should contain all composites that were already in $J$. However, I do understand that generally $F(J)$ may not contain all composites in $C$ so it may not be a subcategory of $C$.

   In the example I gave where $J$ is given by $a\stackrel{f}{\to} b$, the image $F(J)$ contains all composites that were already in $J$, but it does not contain all composites in $C$ when we set $F(a) = F(b)$ so that $F(f)$ is a loop, because $F(f)\circ F(f)$ is not in the diagram in $C$ so the diagram is not a subcategory.

   Instead, you’re usually drawing the image of a map from a quiver.

   In the case of

   $$a \stackrel{f}{\to} b\stackrel{g}{\to} c$$

   that is exactly what is going on.

   Yeah, I understand this. My question was pretty clumsy, but I was thinking in terms of functors. The above is not the image of some $J$ for any functor because it doesn’t contain the composite $g\circ f:a\to c$. So in a way, it is “not allowed”. But as a quiver, I understand it is perfectly legit and the free category generated from this would include the composite.

   And the great thing about diagrams from quivers is that you can take the same concrete definition of limit and apply it directly to them, without bothering about the missing composites and identity morphisms; they won’t change the result!

   Very neat :) So yeah, taking limits of arbitrary quivers will provide all kinds of exercises :)

   Is this because of the adjoint relation? When $F:C\to D$ and $G:D\to C$ are adjoint, do $F$ and $G$ preserve all finite limits?

   Oh, I see the theorem you are probably referring to :)

   adjoint functor theorem

   So the right adjoint preserves limits (which I guess the free functor would be right adjoint) and the left adjoint (which would be the forgetful functor) preserves colimits.

   Thanks :)

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeJul 26th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>Since a diagram is a functor, the image $F(J)$ should contain all composites that were already in $J$. However, I do understand that generally $F(J)$ may not contain all composites in $C$ so it may not be a subcategory of $C$. Yes, exactly. >The above is not the image of some $J$ for any functor because it doesn't contain the composite $g\circ f:a\to c$. So in a way, it is "not allowed". Right. But notice that people very much like to draw diagrams just like the ones that you drew. Even people who, when you ask them to define the term ‘diagram’, will define it to be simply a functor and not breathe a word about quivers or graphs, will still actually draw on paper something that is "not allowed". So you have to figure out for yourself what the original $J$ was and what parts of the diagram are not being drawn (but which can be recovered as composites of what was drawn). >I guess the free functor would be right adjoint No, that's the other way around. See [[free functor]]; we never call anything ‘free’ unless it is (or is an object given by) a *left* adjoint. The adjunction between quivers and strict categories goes between $Quiv$ and $Cat$, but we are not taking any limits (or colimits) in $Quiv$ or $Cat$. Instead, we are taking limits in $C$, where $C$ is some particular *object* of $Cat$. So as far as I can tell, that $G\colon Cat \to Quiv$ preserves all limits, while true, is not relevant here. (I wouldn't be surprised if there was some high-brow abstract way to look at limits that *does* make it relevant, but if so, I don't see it.)

   Since a diagram is a functor, the image $F(J)$ should contain all composites that were already in $J$. However, I do understand that generally $F(J)$ may not contain all composites in $C$ so it may not be a subcategory of $C$.

   Yes, exactly.

   The above is not the image of some $J$ for any functor because it doesn’t contain the composite $g\circ f:a\to c$. So in a way, it is “not allowed”.

   Right. But notice that people very much like to draw diagrams just like the ones that you drew. Even people who, when you ask them to define the term ‘diagram’, will define it to be simply a functor and not breathe a word about quivers or graphs, will still actually draw on paper something that is “not allowed”. So you have to figure out for yourself what the original $J$ was and what parts of the diagram are not being drawn (but which can be recovered as composites of what was drawn).

   I guess the free functor would be right adjoint

   No, that’s the other way around. See free functor; we never call anything ‘free’ unless it is (or is an object given by) a left adjoint.

   The adjunction between quivers and strict categories goes between $Quiv$ and $Cat$, but we are not taking any limits (or colimits) in $Quiv$ or $Cat$. Instead, we are taking limits in $C$, where $C$ is some particular object of $Cat$. So as far as I can tell, that $G\colon Cat \to Quiv$ preserves all limits, while true, is not relevant here. (I wouldn’t be surprised if there was some high-brow abstract way to look at limits that does make it relevant, but if so, I don’t see it.)

5. • CommentRowNumber5.
   • CommentAuthorEric
   • CommentTimeJul 26th 2010
   • (edited Jul 26th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItex>Yes, exactly. It is so rare I hear those words, I think I'll take a moment and savor that :) >No, that's the other way around. I knew I was on thin ice with that one. I see what you mean. I'll put this idea on the back burner and hope to come back to it some day. It might be interesting to think about some more to see if there is any connection. You can take limits in $Cat$, but then you can also take limits in the objects of $Cat$ (obviously), so it would be interesting to see if you could relate the two different levels of limits somehow. It seems like a stretch (even to me!) but you never know. Back burner :) Thanks

   Yes, exactly.

   It is so rare I hear those words, I think I’ll take a moment and savor that :)

   No, that’s the other way around.

   I knew I was on thin ice with that one. I see what you mean. I’ll put this idea on the back burner and hope to come back to it some day. It might be interesting to think about some more to see if there is any connection.

   You can take limits in $Cat$, but then you can also take limits in the objects of $Cat$ (obviously), so it would be interesting to see if you could relate the two different levels of limits somehow. It seems like a stretch (even to me!) but you never know.

   Back burner :)

   Thanks

6. • CommentRowNumber6.
   • CommentAuthorEric
   • CommentTimeJul 26th 2010
   • (edited Jul 26th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItex>We see this with $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$ >where you have a quiver with $3$ vertices and $3$ edges, which generates a category with $3$ objects and $3 \omega$ morphisms. In $Fin Set$, its limit is $$ \{ x\colon a, y\colon b, z\colon c \;|\; y = f(x), z = g(y), x = h(z) \} ,$$ Also, since I like to think of a limit as a "commutizer", i.e. it forces all paths to commute, so starting at any object, tracing around the loop commutes with the identity and we have $h\circ g\circ f\circ x = x$ with cyclic relations $g\circ f\circ h\circ z = z$ and $f\circ h\circ g\circ y = y$ where I guess you are using the [[generalized elements]] $L\stackrel{x}{\to} a$, $L\stackrel{y}{\to} b$, and $L\stackrel{z}{\to} c$, right?

   We see this with

   $$\begin{aligned} &&a&& \\ &\swarrow&&\nwarrow& \\ b&&\rightarrow && c \end{aligned}$$

   where you have a quiver with $3$ vertices and $3$ edges, which generates a category with $3$ objects and $3 \omega$ morphisms. In $Fin Set$, its limit is

   $$\{ x\colon a, y\colon b, z\colon c \;|\; y = f(x), z = g(y), x = h(z) \} ,$$

   Also, since I like to think of a limit as a “commutizer”, i.e. it forces all paths to commute, so starting at any object, tracing around the loop commutes with the identity and we have

   $h\circ g\circ f\circ x = x$

   with cyclic relations

   $g\circ f\circ h\circ z = z$

   and

   $f\circ h\circ g\circ y = y$

   where I guess you are using the generalized elements $L\stackrel{x}{\to} a$, $L\stackrel{y}{\to} b$, and $L\stackrel{z}{\to} c$, right?

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeJul 26th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>I guess you are using the generalized elements Well, *I* wasn't using generalised elements, since I was working in $Fin Set$, where honest to goodness elements really exist. There the limit $L$ is a set whose literal elements are triples $(x,y,z)$, where $x$ is a literal element of $a$, $y$ is a literal element of $b$, and $z$ is a literal element of $c$, such that $f(x) = y$, $g(y) = z$, and $h(z) = x$. However, it's good that you bring up generalised elements, because that is exactly how you can take that construction in $Fin Set$ and reinterpret it in any category $C$ whatsoever. I expect that you were using $L$ to stand for ‘limit’, and indeed the limit does come equipped with generalised elements $x$, $y$, and $z$ as you drew, satisfying the relations $f \circ x = y$, $g \circ y = z$, and $h \circ z = x$. But you can also take *any* object $O$ in the category $C$ and say that the $O$-shaped generalised elements of the limit $L$ (that is, the morphisms from $O$ to $L$) correspond precisely to the triples $(x,y,z)$, where $x$ is an $O$-shaped element $a$, $y$ is an $O$-shaped element of $b$, and $z$ is an $O$-shaped element of $c$, such that $f \circ x = y$, $g \circ y = z$, and $h \circ z = x$. (The relations $h \circ g \circ f \circ x = x$ and its companions follow from these, but they are not enough by themselves.) So if you interpret set-builder notation in this way, then you can write $\{ x\colon a, y\colon b, z\colon c \;|\; y = f(x), z = g(y), x = h(z) \}$ in *any* category $C$ to stand for the limit of that diagram. Then the *specific* triple $(x,y,z)$ of generalised elements that you wrote down corresponds, in this way, to a specific $L$-shaped element of $L$, that is to a specific morphism from $L$ to $L$: the identity morphism on $L$.

   I guess you are using the generalized elements

   Well, I wasn’t using generalised elements, since I was working in $Fin Set$, where honest to goodness elements really exist. There the limit $L$ is a set whose literal elements are triples $(x,y,z)$, where $x$ is a literal element of $a$, $y$ is a literal element of $b$, and $z$ is a literal element of $c$, such that $f(x) = y$, $g(y) = z$, and $h(z) = x$.

   However, it’s good that you bring up generalised elements, because that is exactly how you can take that construction in $Fin Set$ and reinterpret it in any category $C$ whatsoever.

   I expect that you were using $L$ to stand for ‘limit’, and indeed the limit does come equipped with generalised elements $x$, $y$, and $z$ as you drew, satisfying the relations $f \circ x = y$, $g \circ y = z$, and $h \circ z = x$.

   But you can also take any object $O$ in the category $C$ and say that the $O$-shaped generalised elements of the limit $L$ (that is, the morphisms from $O$ to $L$) correspond precisely to the triples $(x,y,z)$, where $x$ is an $O$-shaped element $a$, $y$ is an $O$-shaped element of $b$, and $z$ is an $O$-shaped element of $c$, such that $f \circ x = y$, $g \circ y = z$, and $h \circ z = x$. (The relations $h \circ g \circ f \circ x = x$ and its companions follow from these, but they are not enough by themselves.)

   So if you interpret set-builder notation in this way, then you can write $\{ x\colon a, y\colon b, z\colon c \;|\; y = f(x), z = g(y), x = h(z) \}$ in any category $C$ to stand for the limit of that diagram.

   Then the specific triple $(x,y,z)$ of generalised elements that you wrote down corresponds, in this way, to a specific $L$-shaped element of $L$, that is to a specific morphism from $L$ to $L$: the identity morphism on $L$.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJul 26th 2010
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex> You can take limits in Cat, but then you can also take limits in the objects of Cat (obviously), so it would be interesting to see if you could relate the two different levels of limits somehow. It seems like a stretch (even to me!) but you never know. Actually, you can! There are two things you can say along these lines: * If $F: J \to Cat$ is a diagram in Cat and $f:j \to laxlim F$ is a diagram in the lax limit of F (itself a category), and if the each composite with the projection $p_x \circ f \colon j \to F(x)$ has a limit in the categories $F(x)$, then these limits induce an object of $laxlim(F)$ which is a limit of $f$. Dually, colimits lift to colax limits of categories. * In the same situation, if we consider the *pseudo* limit, rather than the lax limit, we can reach the same conclusion as long as we assume additionally that the limits of the $p_x \circ f$ are preserved by the transition functors $F(\alpha)$, for any arrow $\alpha$ in $J$. One obvious special case of this is that a limit in a product of categories $C\times D$ is just a pair of limits, one in C and one in D. A much more interesting and useful special case of the first fact is that the [[Eilenberg-Moore category]] of a monad is a lax limit, and therefore it inherits all limits from the base category of the monad. It is not an colax limit, so it doesn't inherit all colimits, but the second fact says that it does inherit whatever colimits are preserved by the monad. All of these facts are well-known, but the general statement about 2-categorical limits doesn't seem to be very well known. At least, I've never found it written down anywhere.

   You can take limits in Cat, but then you can also take limits in the objects of Cat (obviously), so it would be interesting to see if you could relate the two different levels of limits somehow. It seems like a stretch (even to me!) but you never know.

   Actually, you can! There are two things you can say along these lines:

   • If $F: J \to Cat$ is a diagram in Cat and $f:j \to laxlim F$ is a diagram in the lax limit of F (itself a category), and if the each composite with the projection $p_x \circ f \colon j \to F(x)$ has a limit in the categories $F(x)$, then these limits induce an object of $laxlim(F)$ which is a limit of $f$. Dually, colimits lift to colax limits of categories.

   • In the same situation, if we consider the pseudo limit, rather than the lax limit, we can reach the same conclusion as long as we assume additionally that the limits of the $p_x \circ f$ are preserved by the transition functors $F(\alpha)$, for any arrow $\alpha$ in $J$.

   One obvious special case of this is that a limit in a product of categories $C\times D$ is just a pair of limits, one in C and one in D. A much more interesting and useful special case of the first fact is that the Eilenberg-Moore category of a monad is a lax limit, and therefore it inherits all limits from the base category of the monad. It is not an colax limit, so it doesn’t inherit all colimits, but the second fact says that it does inherit whatever colimits are preserved by the monad. All of these facts are well-known, but the general statement about 2-categorical limits doesn’t seem to be very well known. At least, I’ve never found it written down anywhere.

9. • CommentRowNumber9.
   • CommentAuthorEric
   • CommentTimeJul 27th 2010
   • (edited Jul 27th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexIs there a reference for lax limit on the nLab? Edit: Google to the rescue :) [[2-limit#lax|2-limit]]

   Is there a reference for lax limit on the nLab?

   Edit: Google to the rescue :)

   2-limit

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeJul 27th 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexRemember, Eric, whenever you want to link to a section within an nLab article, you should give that section a name, so that your link will continue to work even if the sections within the page are rearranged. So edit [[2-limit]] so that the section ## Lax limits becomes ## Lax limits {#lax} Then the link [[2-limit#lax]] will always work.

    Remember, Eric, whenever you want to link to a section within an nLab article, you should give that section a name, so that your link will continue to work even if the sections within the page are rearranged.

    So edit 2-limit so that the section

    ## Lax limits
    

    becomes

    ## Lax limits 
    {#lax}
    

    Then the link 2-limit#lax will always work.

11. • CommentRowNumber11.
    • CommentAuthorEric
    • CommentTimeJul 27th 2010
    • PermaLink
    Author: Eric
    Format: MarkdownItex>Remember, Eric I can't remember something I never knew, but now that I know, I will try to remember :) That is neat. Thanks for the tip. I changed the link above.

    Remember, Eric

    I can’t remember something I never knew, but now that I know, I will try to remember :) That is neat. Thanks for the tip. I changed the link above.