# Isomorphisms in terms of pullbacks

Given functions $f : X \rightarrow Y$ and $g : Y \rightarrow X,$ we say that $f$ and $g$ are inverses iff the following holds.

$$fx = y \Leftrightarrow x=gy$$

We can rehash this condition in terms of pullbacks. Given functions $p,q: A,B \rightarrow C,$ define that their pullback is $$p \times_C q = \{(a,b) \in A \times B \mid pa = qb\},$$

in which case it follows that any two functions $f : X \rightarrow Y$ and $g : Y \rightarrow X$ are inverses iff $$(*)\quad f \times_Y \mathrm{id}_Y = \mathrm{id}_X \times_X g.$$

Now as Berci points out, the above pullbacks exist in any category.

Question: Is there a way to phrase $(*)$ so that it makes sense in arbitrary categories?

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Well, the mentioned pullbacks do exist in any category, and they are isomorphic to $X$ and $Y$, respectively:

$f\times_Y 1_Y\,\cong X\ \$ and $\ \ 1_X\times_X g\,\cong Y$,

so, we indeed have a statement in general: $f\times_Y1_Y\cong 1_X\times_Xg\ \iff\ X\cong Y$, though it sounds somewhat trivial.

(In full generality I'm afraid we cannot expect more, as pullbacks are unique only up to isomorphism, so we can hardly interpret strict equations.)

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How do we know they exist in any category? –  goblin Feb 12 '14 at 23:32
Given any arrow $f:X\to Y$, then draw up the first square you could draw up that has $f$ on the right and $1_Y$ on the bottom and $X$ is the top left object. Verify that this square satisfies the pullback property. –  Berci Feb 12 '14 at 23:37
Well there's an obvious square, but why must it have the pullback property? –  goblin Feb 12 '14 at 23:40
@user18921, in this square $P$ would be $X$ so you would take $u$ to be the same as $q_1$. –  Santiago Canez Feb 13 '14 at 0:44
@Berci, while the given condition on the pullbacks is equivalent to $X \cong Y$, I think the question asked is slightly different: are $f$ and $g$ themselves the maps realizing this isomorphism? I don't see immediately why this is true, do you claim it is? –  Santiago Canez Feb 13 '14 at 14:54

We can reformulate your pullback a little: $f:X\to Y,g:Y\to X$ are mutual inverses just if $Y$ is a pullback $f\times 1_Y$ with upper legs of the pullback square $g$ and $1_Y$ or, equivalently, if $X$ is a pullback $g\times 1_X$ with upper legs $f$ and $1_X$. This is still pretty vacuous, but at least all the maps in the diagram are determined.

Here's a somewhat analogous situation in which there's a less empty result: a map $f:X\to Y$ is a monomorphism if and only if $X$ is a pullback $f\times_Y f$ with the upper legs of the pullback square both $1_X$. In the same way $f$ is an epimorphism if and only if the dual square is a pushout. So you can get that $f$ is mono and epi by examining two squares, though they're not both pullbacks. In many (but not all!) categories, for instance abelian categories and toposes, that's enough to show $f$ is an isomorphism.

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