I understand the proof of the proposition, but I don't see how the functions in the tensored sequence will end up being $f\otimes 1, g\otimes 1$. In the first step of the proof, the functions change from $f\mapsto \bar{f}, \bar{f}(h)=h\circ f$, but it's unclear how the functions change beyond that. Can someone explain?
In case the image doesn't load, the proposition states the following:
Let
$
M'\xrightarrow[]{f}M\xrightarrow[]{g}M''\xrightarrow[]{}0$
be an exact sequence, and let N be any module. Then,
$ M'\otimes N\xrightarrow[]{f\otimes Id_N}M\otimes N\xrightarrow[]{g\otimes Id_N}M''\otimes N\xrightarrow[]{}0$
is exact.
Best Answer
As you wrote, when we pass from $E$ to $\text{Hom}(E,\text{Hom}(N,P))$, the homomorphisms become \begin{equation} 0 \longrightarrow \text{Hom}(M'',\text{Hom}(N,P)) \overset{- \circ g}{\longrightarrow }\text{Hom}(M,\text{Hom}(N,P)) \overset{- \circ f}{\longrightarrow} \text{Hom}(M',\text{Hom}(N,P)). \end{equation} We now apply the isomorphism $\text{Hom}(M \otimes N,P) \cong \text{Hom}(M,\text{Hom}(N,P))$ and look at the induced homomorphisms. More precisely, the map $\text{Hom}(M'' \otimes N,P) \to \text{Hom}(M \otimes N,P)$ is obtained by composing \begin{equation}\require{AMScd} \begin{CD} \text{Hom}(M'',\text{Hom}(N,P)) @>{- \circ g}>> \text{Hom}(M,\text{Hom}(N,P))\\ @AAA @VVV \\ \text{Hom}(M'' \otimes N,P) @. \text{Hom}(M \otimes N,P) \end{CD} \end{equation} Since morphisms in $\text{Hom}(M \otimes N,P)$ correspond to bilinear maps $M \times N \to P$, we see that \begin{equation}\require{AMScd} \begin{CD} m'' \mapsto h(m'',-) @>{- \circ g}>> m \mapsto h(g(m),-)\\ @AAA @VVV \\ (m'',n) \mapsto h(m'',n) @. (m,n) \mapsto h(g(m),n) \end{CD} \end{equation} i.e. the map $\text{Hom}(M'' \otimes N,P) \to \text{Hom}(M \otimes N,P)$ is just precomposition with $g \otimes 1$. Similarly, the map $\text{Hom}(M \otimes N,P) \to \text{Hom}(M' \otimes N,P)$ is precomposition with $f \otimes 1$. Going back to the sequence $E \otimes N$, we get \begin{equation} M' \otimes N \overset{f \otimes 1}{\longrightarrow} M \otimes N \overset{g \otimes 1}{\longrightarrow} M'' \otimes N \longrightarrow 0. \end{equation}