# Prove $\text{Hom} (Z, \ker(f, g)) ≃ \ker(\text{Hom} (Z, X) ⇉ \text{Hom} (Z, Y ))$ for sets

I just started reading some Schapira notes on Algebra and Topology. Statement 1.7 is the following:

Hom $(Z, \ker(f, g)) ≃ \ker(\text{Hom} (Z, X) ⇉ \text{Hom} (Z, Y ))$

where:

$Ker(f, g) = \{x ∈ X; f(x) = g(x)\}$

I tried verifying this myself by drawing a diagram, but failed. Can anyone explain why this is true?

-
The way to prove any such isomorphism is to construct a bijection. The way to prove that a map is bijective is to construct its inverse. Can you construct some natural map from LHS to RHS? What about RHS to LHS? – Grigory M Aug 15 '10 at 11:06
Dear Casebash: You may want to write explicitly in your question that the text you link to is from Schapira. – Pierre-Yves Gaillard Aug 16 '10 at 5:15
Thank you for him! – Pierre-Yves Gaillard Aug 16 '10 at 10:10

Maybe something is missing in the original statement and is: whose kernel is that on the right hand side?

First of all, let's recall that $f$ and $g$ are maps from $X$ to $Y$.

Then that kernel on the right is obviously

$$\ker (f_{*}, g_{*}) \subset \mathrm{Hom}(Z,X) \ ,$$

where

$$f_{*}, g_{*} : \mathrm{Hom}(Z,X) \longrightarrow \mathrm{Hom}(Z,Y)$$

are the maps defined by $\varphi \mapsto f\circ \varphi$ and $\varphi \mapsto g\circ \varphi$, respectively.

That said, the bijection you ask for simply sends every map

$$\varphi : Z \longrightarrow \ker (f,g) \subset X$$

to $i\circ \varphi$, where $i: \ker (f,g) \hookrightarrow X$ is the inclusion.

As for the other direction:

$$\psi \in \ker (f_{*}, g_{*}) \ \Longleftrightarrow \ f\circ \psi = g\circ \psi \ \Longleftrightarrow \ f(\psi (z)) = g(\psi (z)) \ \text{for all} \ z \in Z$$

Which means:

$$\psi (z) \in \ker (f,g) \ \text{for all} \ z \in Z \ \Longleftrightarrow \ \mathrm{im} (\psi ) \subset \ker (f, g) \ .$$

So, $\psi : Z \longrightarrow X$ factorises through the inclusion $i: \ker (f,g) \hookrightarrow X$. That is, since $i$ is injective, there exists a unique $\varphi : Z \longrightarrow \ker (f,g)$ such that $\psi = i \circ \varphi$.

Hence, the map $LHS \longrightarrow RHS$ sends $\varphi$ to $i \circ \varphi$ and the map $RHS \longrightarrow LHS$ sends $\psi$ to $\varphi$ such that $\psi = i \circ \varphi$.

Both compositions are the identity by definition:

$$\psi \mapsto \varphi \mapsto i \circ \varphi = \psi$$

and

$$\varphi \mapsto i \circ \varphi \mapsto \varphi \ .$$

Conclusion: I've never seen "two" sets that were so much the same. :-)

-
What does the notation φ↦i∘φ↦φ mean? I've only ever seen that used with a single arrow – Casebash Aug 15 '10 at 21:14
@Casebash. $\varphi \mapsto i\circ \varphi \mapsto \varhi$ means: first you send $\varphi$ to $i \circ \varphi$ with the map from LHS to RHS, then you send $i \circ \varphi$ to $\varphi$ with the map from RHS to LHS. – a.r. Aug 16 '10 at 1:23

This is a special case of the theorem that covariant hom-functors (here $\mathrm{Hom}(Z,\underline{\ })$) preserve limits. This is Theorem 1 in section V.4 of Mac Lane's Categories for the Working Mathematician.

-
Isn'it what you call a sledgehammer? :) – Pierre-Yves Gaillard Aug 15 '10 at 16:41
It may be a sledgehammer, but then again what is the first example of a continuous functor that anyone studying category theory will see? You guessed it, representables. Besides, preservation of limits is (equivalent to) the statement that a certain canonical map is an isomorphism, much stronger than the mere fact that some isomorphism exists. After Understanding this, solving the exercise is only an inch away. And the general case of limit-preservation is not significantly harder than the kernel-preserving one. – G. Rodrigues Aug 16 '10 at 2:10
This was a reference to math.stackexchange.com/questions/2367/… and was kind of a joke. I think Agusti's answer is much easier to read than Mac Lane's proof. The best is of course to understand the wider context, and at the same time to be able to give a no nonsense proof of the statement in the question. --- By the way, I voted for this answer (and also for Agusti's, and for the question (I'm a positive person)). – Pierre-Yves Gaillard Aug 16 '10 at 5:34
My comment was in no way meant to be disparaging, although it is possible that its tone may convey exactly that feeling (english is not my native language and personally, I find it very hard to convey the finer shades of meaning and feeling); and besides, the smiley at the end of your comment made your intentions clear. I think the point I was trying to make is best summed up by quoting you "The best is of course to understand the wider context" as such understanding makes such no-nonsense proofs possible. – G. Rodrigues Aug 16 '10 at 13:31
I didn't find your comments disparaging at all! On the contrary, I found them very interesting! – Pierre-Yves Gaillard Aug 16 '10 at 17:08

I think Hom (Z, X) ⇉ Hom (Z, Y) means some pair from this domain, rather than all possible pairs. If I am wrong, then this solution will need to be changed. If it is a single, pair, then it is pretty clear that it will be f◦, g◦

LHS=$K_1$ is all k:Z→X'=Ker(f,g)$\in$X and f(x)=g(x) for x$\in$X' RHS=$K_2$ is all k:Z→X with f◦(k)=g◦(k) or f◦(k)(z)=g◦(k)(z) for z$\in$Z

Consider $k \in K_1$. As f(x)=g(x),f◦(k)(z)=g◦(k)(z) for all z. So $k \in K_2$ Now consider $k \in K_2$. f◦(k)=g◦(k). If k(z) not in Ker(f,g) for some z, then f◦(k)(z)$\ne$g◦(k)(z), so contradiction. So $k \in K_1$

Therefore, there is a bijection between $k \in K_1$ and $k \in K_2$

-
I think Agusti is right. The only thing to realize is that an element of the second kernel maps $Z$ into $Ker(f,g)$. – Pierre-Yves Gaillard Aug 15 '10 at 13:22