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?
 A: 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.
A: 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.   :-)
A: 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 $k \in K_2$
