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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?

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    Dear Casebash: You may want to write explicitly in your question that the text you link to is from Schapira.2010-08-16

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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 $k \in K_2$

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    I think Agusti is right. The only thing to realize is that an element of the second kernel maps $Z$ into $Ker(f,g)$.2010-08-15
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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.

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    I didn't find your comments disparaging at all! On the contrary, I found them very interesting!2010-08-16
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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. :-)

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    @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.2010-08-16