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The Tensor-Hom adjunction is the natural isomorphismFor an R-module homomorphism $f$, it is given explicitly byConversely, an $R$-linear map $g:M\to {\mathrm{Hom}}_R(N,P)$ determines the balanced map $(m,n)\mapsto g(m)(n)$, so the universal property of the tensor product of modules givesThese formulas are inverse to each other because pure tensors generate the tensor product of modules.

Solved by gpt-5.6-sol high.

Let $u:M^{\prime }\to M$ be an R-module homomorphism. Naturality in the left argument means that precomposition by $u\otimes 1_N$ on the left corresponds under the Tensor-Hom adjunction to precomposition by $u$ on the right. For $f:M{\otimes }_{R}N\to P$,Thus the naturality square commutes pointwise on every $m^{\prime }\in M^{\prime }$ and $n\in N$.

Solved by gpt-5.6-sol high.