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?enter image description here

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.

  • $\begingroup$ To me, it is unclear what you are actually asking. Note also that it is bad policy to supply crucial information in a picture (this may not display properly on mobile devices, it is not searchable, etc.). For this reason, you should paraphrase the contents of said Proposition 2.18 and comment on the parts of the proof that you cannot follow. $\endgroup$
    – Claudius
    Jun 17, 2022 at 13:13
  • $\begingroup$ I have edited the question. I understand the short proof written in the text fine, but it doesn't seem to explain/justify why the mappings in the tensored sequence will be $f\otimes 1, g\otimes 1$. When I try to reproduce the steps of the proof explicitly, I end up with some as-yet undetermined functions $f', g'$ in the tensored exact sequence. $\endgroup$
    – A.D.
    Jun 17, 2022 at 13:19
  • $\begingroup$ But this is a straightforward computation. Could you explain to us which part of this computation you have trouble with? Surely there must have been some things that you have tried or that confuse you. $\endgroup$
    – Claudius
    Jun 17, 2022 at 13:31
  • $\begingroup$ We first conclude that the sequence $0\rightarrow Hom(M'', Hom(N, P))\xrightarrow[]{\bar{g}}Hom(M, Hom(N, P))\xrightarrow[]{\bar{f}}Hom(M', Hom(N, P))$ is exact. This is fine. The next sequence is $0\rightarrow Hom(M''\otimes N, P)\xrightarrow[]{\bar{g}'}Hom(M\otimes N, P)\xrightarrow[]{\bar{f}'}Hom(M'\otimes N, P)$. I'm getting stuck in the movement from $\bar{f}$ to $\bar{f'}$. We know that $\bar{f}(\phi)=\phi\circ f$, where $\phi: M\rightarrow Hom(N, P)$. How do you change from this to a function which acts on $\phi'\in Hom(M\otimes N, P)$ in an equivalent manner? $\endgroup$
    – A.D.
    Jun 17, 2022 at 13:38
  • 1
    $\begingroup$ Well, let $\alpha\colon \operatorname{Hom}(M,\operatorname{Hom}(N,P)) \to \operatorname{Hom}(M\otimes N,P)$ be the isomorphism in (1), given explicitly by $\alpha(\varphi)(m\otimes n) = \varphi(m)(n)$. The inverse map $\beta$ is given by $\beta(\psi)(m)(n) = \psi(m\otimes n)$. Then $\bar{f}' = \alpha\circ \bar{f} \circ \beta$ by definition. Since you know what $\bar{f}$ is, you should be able to compute $\bar{f}'$ from this. $\endgroup$
    – Claudius
    Jun 17, 2022 at 13:46

1 Answer 1


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}


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