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Wikipedia presents composition as a special case of pullbacks, but I can't quite reconcile that interpretation with the definition of pullback that I know.

In its most general form, composition refers to three morphisms and three objects, whereas the definition I know of pullback is stated in terms of four objects and four morphisms, although, if we are claiming that the composition is a special case of pullback, then some of the morphisms in this "special-case pullback" could be identities (so their domains and codomains would be equal).

Following up on this idea,

Of course, if I have morphisms $f:X\to Y$, $g:Y\to Z$ and $h:X\to Z$ such that $f\;;g = h$, (where I'm using "$f\;;g\;$" as just another way to write "$g\circ f\;$"), then the following square diagram commutes:

$$ \begin{array}{cccc} &X & \stackrel{f}{\to} & Y\\ {\small 1_X}\!\!\!\!\!\!\!\!\!&\downarrow & &\downarrow {\small g}\\ &X & \stackrel{h}{\to} & Z \end{array} $$

Therefore it meets the first condition for being the pullback of $h$ and $g$, but it seems to me that it fails to meet the second condition (universality). I give my proof of this below, FWIW, but my question is:

Is there some other way to express composition as a pullback? Alternatively, is there some unstated assumption in the Wikipedia description?

To prove that the diagram above, in general, does not have the universality property of a pullback, here's a counterexample in Set. Using the diagram above, let $X = Y = \mathbf{2}$, $Z = \mathbf{1}$, $f = 1_{\mathbf{2}}$, and $h = g = \;\;!$, the unique morphism $\mathbf{2}\to \mathbf{1}$. Suppose, then, that this is a pullback square:

$$ \begin{array}{cccc} &\mathbf{2} & \stackrel{1_{\mathbf{2}}}{\to} & \mathbf{2}\\ {\small 1_{\mathbf{2}}}\!\!\!\!\!\!\!\!\!&\downarrow & &\downarrow {\small !}\\ &\mathbf{2} & \stackrel{!}{\to} & \mathbf{1} \end{array} $$

Now, setting $W = \mathbf{2}$, $p:W\to X$ and $q:W\to Y$ respectively equal to $1_{\mathbf{2}}$ and $\sigma \neq 1_{\mathbf{2}}$ (e.g. $\sigma$ could be the odd permutation on $\mathbf{2}$), we have $1_{\mathbf{2}};! = p;h = q;g = \sigma;!$. Therefore, by the universality property of pullbacks, there must exist a unique morphism $u:\mathbf{2} \to \mathbf{2}$ such that $u;1_X = u;1_{\mathbf{2}} = 1_{\mathbf{2}} = p$ and $u;f = u;1_{\mathbf{2}} = \sigma = q$. But if such $u$ existed, we would have $1_{\mathbf{2}} = \sigma$, which is a contradiction.

Edit: by way of clarification, I want to point out that the correct pullback (as I understand the concept) for the pair of morphisms $(!, !)$ shown in the counterexample above is the product $\mathbf{2}\times\mathbf{2}$, together with the standard projections $\pi_0, \pi_1:\mathbf{2}\times\mathbf{2}\twoheadrightarrow\mathbf{2}$. In fact, at least in Set, the pullback of two morphisms $A\twoheadrightarrow\mathbf{1}$ and $B\twoheadrightarrow\mathbf{1}$ is the only time that the pullback takes up all of $A\times B$, in which case, the universality property of the pullback coincides with the universality property of products.

One more clarification, the point of my counterexample is to show that the naive way of getting a pullback diagram out of a composition diagram that I showed at the beginning of this post (namely, by adding an identity edge to the composition diagram) cannot be correct, so the Wikipedia description must be referring to something else. The aim of my question is to determine what this "something else" is.

Lastly, this other way of getting a pullback diagram from a composition

$$ \begin{array}{cccc} &X & \stackrel{f}{\to} & Y\\ {\small h}\!\!\!\!\!\!\!\!\!&\downarrow & &\downarrow {\small g}\\ &Z & \stackrel{1_Z}{\to} & Z \end{array} $$

is wrong too: since $1_Z$ is monic, the diagram can be a pullback square only if $f$ is monic too, which certainly fails to hold in general for a composition $f\;;g = h$.

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I believe Wikipedia is using pullback in the looser sense of the result of a contravariant functor acting on an arrow. In this case, the contravariant functor in question is the contravariant Hom functor $\textrm{Hom}(-, Z)$. –  Zhen Lin Nov 12 '11 at 18:02
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The remainder of the article seems to explain how these are distinct but related. Did that help at all? –  Dylan Moreland Nov 12 '11 at 20:36
    
I assume that you've seen the first line of that Wikipedia entry: "a pullback is either of two different, but related processes: precomposition and fiber-product." –  t.b. Nov 13 '11 at 13:21
    
@ZhenLin: I will accept your answer if you post it. –  kjo Nov 13 '11 at 13:27
    
@tp1: thank you! –  kjo Nov 13 '11 at 14:56
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2 Answers 2

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A little more detail on my comment.

There is a general notion of ‘pullback’ used informally throughout mathematics to refer to, well, pulling back some structure or other through a given map. Category-theoretically speaking, ‘pulling back through/along $f : A \to B$’ usually corresponds to a map $F f : FB \to FA$ where $F$ is some contravariant functor. Explicitly:

  1. The inverse image map, a.k.a. ‘pulling a subset of $B$ back along $f$’, is the map $f^* : \mathscr{P} B \to \mathscr{P} A$ where $\mathscr{P} : \textbf{Set} \to \textbf{Set}^\textrm{op}$ is the (contravariant) power set functor.

  2. Generalising slightly, pulling back along an arrow $f : A \to B$ in a category $\mathbf{C}$ with all pullbacks induces a functor $f^* : \mathbf{C} / B \to \mathbf{C} / A$, where $\mathbf{C} / A$ denotes the slice category over $A$.

  3. Specialising the above example, if we take $\mathbf{C} = \textbf{Top}$, then we get the notion of pullback bundles. For example, if $TN \to N$ is the tangent bundle over a manifold $N$, and $f : M \to N$ is smooth, then we have a bundle $f^* TN \to M$; the differential of $f$ then becomes a map $df : TM \to f^* TN$ of bundles over $M$.

  4. Finally, consider the functor $\textrm{Hom}(- , Z)$. If we have a map $f : A \to B$, then we get the precomposition map $f^* : \textrm{Hom}(B, Z) \to \textrm{Hom}(A, Z)$.

  5. Returning to the example of bundles, consider the space of global sections $\Gamma(N, T^* N)$ of the cotangent bundle $T^* N \to N$. This is a subset of $\textrm{Hom}(N, T^* N)$, so precomposing with $f$ yields a subset of $\textrm{Hom}(M, T^* N)$. But these factor through $f^* T^*N$, so we get a subspace of $\Gamma (M, f^* T^* N)$. And $f^* T^* N$ is simply the dual bundle of $f^* TN$ so we have a dual map $df^* : f^* T^*N \to T^* M$, so postcomposing with this yields a subspace of $\Gamma(M, T^* M)$. This map $\Gamma(N, T^* N) \to \Gamma(M, T^* M)$ is what geometers mean by ‘pullback of differential forms’, and as you can see, there are lots of contravariant functors involved!

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Thanks for this explanation. The inverse image map example helped very much. –  tp1 Nov 13 '11 at 19:08
    
@Zhen Lin: Thanks! There's a whole lot of math in that reply! If I'm not mistaken, #2 corresponds to the pullback I'm familiar with. #3 is at the edge of what I can follow, and #5 is definitely beyond it (but it gives me something to shoot for). All your examples, even the ones above my head, really show how the "categorical pullback" generalizes the pullbacks found in other areas of math. (If I understood the contents of the Wikipedia Pullbacks page better, I'd offer to port your write-up to it, but as of now, I'd be the proverbial bull in the china shop...) –  kjo Nov 13 '11 at 22:09
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As far as I understand you want to check that if you have an object $W$ and two morphisms $q:W \rightarrow Y, t:W \rightarrow X$ such that $gq=gft$ then there exists a unique morphism $w:W \rightarrow X$ such that $fw=q, 1_{W}w=t,$ take $w=t$ than it sattisfies the conditions and is unique ($1_{W}w'=t$ implies $w'=t$).

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The point of my counterexample was to show a case in which that morphism $w:W\to X$ does not exist. I agree with you that the fact that one of the square's sides is $1_X$ (not $1_W$ as you had originally), or, more generally, the fact it is monic, necessarily makes such $w$ unique, if it exists. Existence is not guaranteed, as my counterexample shows. When the morphism $g$ is also monic, the problem becomes overconstrained. The pullback for the $!, !:\mathbf{2}\to\mathbf{1}$ pair is $\mathbf{2}\times\mathbf{2}$, with the standard projections onto $\mathbf{2}$. –  kjo Nov 13 '11 at 12:27
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