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$\DeclareMathOperator\Hom{Hom}$I have the following exercise in my class of Category Theory:

Prove that $\text{Hom}(Z,\Hom(Y,X))\cong \Hom(Y*Z, X)$ but I am not sure what $*$ is. I think that $*$ should be $+$ (by $+$ I mean the co-product). Because at $\operatorname{Hom}(X \times Z, Y) \cong \operatorname{Hom}(X, \operatorname{Map}(Z, Y))$ is not true in $\textbf{Top}$ it was discussed that $\Hom(Z,\Hom(Y,X))\not \cong \Hom(Y\times Z, X)$.

But I have been trying to find a bijection between $\Hom(Z,\Hom(Y,X))$ and $\Hom(Y+Z, X)$.

I have a map going from $\Hom(Z,\Hom(Y,X))$ to $\Hom(Y+Z, X)$: First of all note that $\Hom(Y+Z, X)$ is in bijection with $\Hom(Y,X)\times\Hom(Z,X). This is very clear; see for example Hom Functor Preserves Coproducts.

So I need to find for each $f:Z\rightarrow \Hom(Y,X)$ maps $Y\rightarrow X$ and $Z\rightarrow X$. To get them I do the following: for each $y\in Y$ and $z\in Z$ I define $$Y\to X; y\mapsto z(y)$$ and $$Z\to X;z\mapsto z(y).$$

But I don't know how to get arrows in the other direction.

Any help would be helpful.

LSpice
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Gillyweeds
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    $\operatorname{Hom}(Z, \operatorname{Hom}(Y, X))$ does not make sense in an abstract category $\mathcal C$, since $\operatorname{Hom}(Y, X)$ is not a priori an element of $\mathcal C$. (Note that your linked question has to impose a topology—the compact-open topology—on $\operatorname{Map}(Y, X)$.) What is the additional structure that allows us to make sense of this nested Hom-set? – LSpice Mar 12 '22 at 19:06
  • What if we are in the set category? – Gillyweeds Mar 12 '22 at 19:10
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    Then this is a question in set theory, not category theory, and you can check that $(X^Y)^Z \cong X^{Y \times Z}$ via the obvious (un)currying map. – LSpice Mar 12 '22 at 19:12
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    Okay thank you very much! – Gillyweeds Mar 12 '22 at 19:18
  • If we had $(X^Y)^Z \cong X^{Y+Z}$ instead of $X^{Y\times Z}$, then $A^B$ for the set of functions $B\to A$ would be a very bad choice of notation! – Alex Kruckman Mar 13 '22 at 00:33

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