# Kernels in $\mathbf{Top}$

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?

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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

In universal algebra this concept is also known as the kernel of a morphism $f : X\to Y$. However, topological spaces are not universal algebras and calling these things "kernels" is not a good idea, because the term is reserved for a better known concept already.

Incidentally, in a typical "concrete category of sets with some structure", like $\mathsf{Top}$, this kind of "kernel" is a particular case of a pullback $(P,p_1 : P \to X_1, p_2 : P \to X_2)$ of morphisms $f_1 : X_1 \to Y$ and $f_2 : X_2\to Y$ with common codomain. It can be constructed via:

$$P = \{ (x_1,x_2)\in X_1\times X_2 : f_1(x_1) = f_2(x_2)\}$$

and choosing $p_1,p_2$ to be canonical projections. It should be clear, that you get this "kernel" by choosing $f = f_1 = f_2$.

A kernel pair of a morphism $f$ in a category is the pullback of $f$ along itself. This is standard modern terminology for the notion you are interested in.

Is this notion really important in a more general context? Yes, definitely! In a category with finite limits (that is: pullbacks and a terminal object) (such as $\mathsf{Top}$ or every algebraic category) a kernel pair is an internal equivalence relation. Yes, these are exactly what you think they are (in $\mathsf{Set}$ there are the equivalence relations, in algebraic categories they are congruence relations). And quotients by internal equivalence relations are taken by taking their coequalizer. It all works out nicely.

One might also be wondering about the relationship between kernel pairs and kernels (as in something like $\{x : f(x) = 1\}$), because in group theory for example they are more commonly used, when talking about quotients (there are exactly the normal subgroups). Kernels are actually examples of equalizers, that is: in a category with a zero object $0$ a kernel of a morphism $f$ is an equalizer of $f$ and the composite $A\to 0 \to B$. Hence, if you know that something is a kernel, then there is always a morphism $f$ and if your category has finite limits, then $f$'s kernel pair is exactly the internal equivalence relation associated with that kernel (e.g. in group theory the relation $x \sim y \Leftrightarrow xy^{-1} \in \ker f$ is really just the kernel pair of $f$).

Remark: Existence of pullbacks suffices to prove, that every kernel pair is an internal equivalence relation, but then the definition linked needs to be adjusted.

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Only one comment : what do you mean precisely by "algebraic" category? (I know a fair amount about categories and algebraic geometry, I've just never heard the term.) Is it just a general idea of the notion or it's some precise technical term? Feel free to be specific or refer to some document. – Patrick Da Silva Jun 28 at 21:46
@PatrickDaSilva It was just supposed to be the general idea. It could mean for example: "category of models of a Lawvere theory in $\mathsf{Set}$". – Stefan Perko Jun 29 at 7:13
@PatrickDaSilva Does this answer suffice? – Stefan Perko Jul 1 at 10:57