Simple question about exponentials I am studying exponentials from MacLane-Moerdijk's book, "Sheaves in geometry and Logic". I do not understand the following: Induced by the product-exponent adjunction, consider the bijection $$\hom(Y\times X,Z)\to\hom(Y,Z^X)\;\;\;\;\;\;(\star)$$
They say: The existence of the above adjunction can be stated in elementary terms (i.e., without using Hom-sets). For, set $Y=Z^X$ in $(\star)$; The identity arrow $1:Z^X\to Z^X$ on the right in $(\star)$ then corresponds, under the adjunction, to an arrow $e:Z^X\times X\to Z$. The bijection $f\mapsto f'$ of $(\star)$, by naturality, now becomes the statement that to each $f:Y\times X\to Z$ there is a unique $f':Y\to Z^X$ such that the diagram 

commutes.
I do not understand how the definition of naturality gives the above statement. Naturality, if I am not mistaken, says (naturality in $Y$, if this is what is meant) that if $$\alpha_Y:\hom(Y\times X,Z)\to\hom(Y,Z^X) $$ then for all $g:Y\to W$,$$(-\circ g)\circ \alpha_W=\alpha_Y\circ(-\circ (g\times1)) $$
How does that give the statement? I am sure I am doing something wrong. Can anyone clarify, please?
 A: The naturality of $\alpha_Y$ in $Y$ means that: for any $Y$, for any $W$, for any $g : Y \rightarrow W$, for any $h: W \times X \rightarrow Z$, we have $\alpha_{W}(h) \circ g = \alpha_Y(h \circ (g \times id_X))$, hence ${\alpha_{Y}}^{-1}(\alpha_{W}(h) \circ g) = h \circ (g \times id_X)$. With $W = Z^X$ and $h = {\alpha_{Z^X}}^{-1}(id_{Z^X}) = e$, we obtain ${\alpha_{Y}}^{-1}(g) = e \circ (g \times id_X)$. 
I show the existence of $f'$: let $f: Y \times X \rightarrow Z$; we set $g = \alpha_Y(f) = f'$.
I show the unicity of $f'$: let $f'_1, f'_2: Y \rightarrow Z^X$ such that $e \circ (f'_1 \times id_X) = e \circ (f'_2 \times id_X)$; we have ${\alpha_Y}^{-1}(f'_1) = {\alpha_Y}^{-1}(f'_2)$, hence $f'_1 = f'_2$.
A: Ok, I believe I have understood this. Can someone check if this is a correct way of viewing it?
Let $\phi_Y \colon \hom(Y, Z^X) \to \hom(Y \times X, Z)$ be the bijection. The identity $1 \colon Z^X \to Z^X$ goes to an arrow $e \colon Z^X \times X \to Z$. By naturality, for any $f' \colon Y \to Z^X$ the following diagram:
$$
  \require{AMScd}
  \begin{CD}
    \hom(Z^X, Z^X) @>{\phi_{Z^X}}>> \hom(Z^X \times X, Z) \\
    @VVV @VVV \\
    \hom(Y, Z^X) @>{\phi_Y}>> \hom(Y \times X, Z)
  \end{CD}
$$
commutes. Particularly, we have:
$$
  \require{AMScd}
  \begin{CD}
    1_{Z^X} @>{\phi_{Z^X}}>> e \\
    @VVV @VVV \\
    f' @>{\phi_Y}>> \phi_Y(f') = e\circ(f' \times 1)
  \end{CD}
$$
But, since $\phi_Y$ is a bijection, each $f \colon Y \times X \to Z$ has the form $f = e \circ (f' \times 1)$ for a unique $f'$, as required.
