Remark 4.43.12. Let $\mathcal{C}$ be a monoidal category. We say $\mathcal{C}$ has an *internal hom* if for every pair of objects $X, Y$ of $\mathcal{C}$ there is an object $hom(X, Y)$ of $\mathcal{C}$ such that we have

functorially in $X, Y, Z$. By the Yoneda lemma the bifunctor $(X, Y) \mapsto hom(X, Y)$ is determined up to unique isomorphism if it exists. Given an internal hom we obtain canonical maps

$hom(X, Y) \otimes X \to Y$,

$hom(Y, Z) \otimes hom(X, Y) \to hom(X, Z)$,

$Z \otimes hom(X, Y) \to hom(X, Z \otimes Y)$,

$Y \to hom(X, Y \otimes X)$, and

$hom(Y, Z) \otimes X \to hom(hom(X, Y), Z)$ in case $\mathcal{C}$ is symmetric monoidal.

Namely, the map in (1) is the image of $\text{id}_{hom(X, Y)}$ by $\mathop{\mathrm{Mor}}\nolimits (hom(X, Y), hom(X, Y)) \to \mathop{\mathrm{Mor}}\nolimits (hom(X, Y) \otimes X, Y)$. To construct the map in (2) by the defining property of $hom(X, Z)$ we need to construct a map

and such a map exists since by (1) we have maps $hom(X, Y) \otimes X \to Y$ and $hom(Y, Z) \otimes Y \to Z$. To construct the map in (3) by the defining property of $hom(X, Z \otimes Y)$ we need to construct a map

for which we use $\text{id}_ Z \otimes a$ where $a$ is the map in (1). To construct the map in (4) we note that we already have the map $Y \otimes hom(X, X) \to hom(X, Y \otimes X)$ by (3). Thus it suffices to construct a map $\mathbf{1} \to hom(X, X)$ and for this we take the element in $\mathop{\mathrm{Mor}}\nolimits (\mathbf{1}, hom(X, X))$ corresponding to the canonical isomorphism $\mathbf{1} \otimes X \to X$ in $\mathop{\mathrm{Mor}}\nolimits (\mathbf{1} \otimes X, X)$. Finally, we come to (5). By the universal property of $hom(hom(X, Y), Z)$ it suffices to construct a map

We do this by swapping the last two tensor products using the commutativity constraint and then using the maps $hom(X, Y) \otimes X \to Y$ and $hom(Y, Z) \otimes Y \to Z$.

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