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> [!proposition] Proposition. ([[the tensor product is left-adjoint to hom]])
> Let $R$ be a [[ring|(say, commutative)]] [[ring]] and $N$ be an $R$-[[module]]. The [[tensor functor]] $\_ \otimes_{R} N$ is [[adjoint functor|left-adjoint]] to the [[hom functor]] $\text{Hom}(N, -)$.[^1]
>
> Explicitly, this is the assertion that for all $R$-[[module|modules]] $M,N,P$ there is a [[natural transformation|natural]] [[bijection]] $\text{Hom}\big(M, \text{Hom}(N,P)\big) \simeq \text{Hom} \big( M \otimes _{R} N, P \big) .$
> This is certainly true: recall that the [[universal property]] defining the [[tensor product of modules|tensor product]] allows us to uniquely identify any $R$-[[linear map|linear map]] $M \otimes_{R} N \xrightarrow{} P$ with a [[bilinear map]] $M \times N \xrightarrow{\varphi} P$. Now,
> - $R$-linearity in the second argument of $\varphi$ just means that the map $\varphi(m,-):N \to P$ is linear for all $m \in M$.
> - $R$-linearity additionally in the first argument means that the map $m \mapsto \varphi(m,-)$ from $M$ to $\text{Hom}(N,P)$ is $R$-linear.
>
> Thus $\varphi$ is an element of $\text{Hom}(M, \text{Hom}(N,P))$.
>
>
>
[^1]:Although in general the [[hom functor]] is valued in $\mathsf{Set}$, recall that [[homsets in R-mod are R-modules]], so we may upgrade in this case to a functor $\text{Hom}(N,-):R\text{-}\mathsf{Mod} \to R\text{-}\mathsf{Mod}$.
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#### References
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