Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

There is a following well-known theorem for abelian categories (at least the ones I know, Ab, $R$-mod and so on... not so familiar with categorical language to be honest) which states the following :

If $X,Y,Z$ are objects and $f : X \to Y$, $g : X \to Z$ morphisms with $g$ surjective, then there exists a unique morphism $h : Z \to Y$ such that the diagram commutes if and only if $\ker g \subseteq \ker f$. (Draw the diagram, won't do it here :D )

Now I was getting started doing algebraic topology in Allen Hatcher's book and some question wanted me to work out homotopies and I realized I implicitly used the following.

If $X,Y,Z$ are topological spaces and $f : X \to Y$, $g : X \to Z$ continuous maps with $g$ being a surjective quotient map (quotient map means $U \subseteq Z$ is open if and only if $g^{-1}(U)$ is open in $X$), then there exists a unique continuous map $h : Z \to Y$ such that the diagram commutes if and only if $$ \ker g \overset{def}= \{ (x_1,x_2) \in X^2 \, | \, g(x_1) = g(x_2) \} \subseteq \{ (x_1,x_2) \in X^2 \, | \, f(x_1) = f(x_2) \} \overset{def}= \ker f. $$ (The map $h$ is obviously defined by $f \circ g^{-1}$ and the condition makes sure that everything works out. I am not sure how to prove that $f \circ g^{-1}$ will be continuous in general or what conditions precisely should be added, I did it in the case of a projection (i.e. $g$ was just a map that "glued points together" and $Z$ was $X / \sim$ for some equivalence relation that glued points). Maybe this is not as general as one could wish. )

My question is : I feel like I didn't totally imagine this notion of kernel in Top, I think I've read it somewhere but I am not sure and I have no idea how to look it up online, since "kernel" redirects to the well-known notion and not to this one. Anyone knows if this kind of kernel is useful in a more general setting, or if some results that are true in Ab also hold in Top with this tweak in a similar way that I did?

share|cite|improve this question
FYI: Neither of those statements hold without some sort of surjectivity condition on $g$. – Jim Aug 13 '13 at 21:39
Surjective + open implies that it is a quotient map, so you're right that it's enough. Stefan is pointing out that you can actually do with less than that. – Jim Aug 13 '13 at 21:45
@Downvoter : What's wrong? Anything I can fix? I think I got it all straightened out... – Patrick Da Silva Aug 13 '13 at 21:58
Well, for any homomorphism of algebras (in the sense of universal algebra), the kernel (defined as in your question) is a congruence relation on the domain of the homomorphism. A homomorphism $f:X\to Y$ factors through a surjection $g:X\to Z$ iff the kernel of $f$ includes that of $g$. Every congruence relation on an algebra is the kernel of a surjective homomorphism (unique up to composition with an isomorphism of the codomain). All this and lots more about kernels should be in any textbook of universal algebra, e.g., the one by Burris and Sankappanavar or the one by Cohn. – Andreas Blass Aug 13 '13 at 23:09
Dear Patrick, If you want to find a discussion of this sort of categorical construction, you could search for phrases like (co)equalizier, strict epimorphism, and universal strict epimorphism. Regards, – Matt E Aug 13 '13 at 23:14

This is known as the kernel pair of a morphism.

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.