Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.)

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.