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Let $M$ be a right $R$-module and $$0 \xrightarrow{} A' \xrightarrow{\phi} A \xrightarrow{\psi} A'' \xrightarrow{} 0$$ is a split short exact sequence of left $R$-modules and $R$-homomorphisms. In the book Module Theory: An Approach to Linear Algebra by T.S.Blyth a proof is given that the induced sequence $$0 \xrightarrow{} M\otimes A' \xrightarrow{1_M\otimes\phi} M\otimes A \xrightarrow{1_M\otimes\psi} M\otimes A'' \xrightarrow{} 0$$ is also split exact. The proof uses the previous proposition which states that the induced sequence is split at $M\otimes A$ and at $M\otimes A''$, and then explicitly constructs a splitting $R$-homomorphism for $\phi$, that is, an $R$-homomorphism $\vartheta\colon M\otimes A \to M\otimes A'$ such that $\vartheta\circ(1_{M}\otimes\phi) = 1_M\otimes A'$.

However, before that it was proved that $(\psi_1\circ\phi_1)\otimes(\psi_2\circ\phi_2) = (\psi_1\otimes\psi_2)\circ(\phi_1\otimes\phi_2)$ and $1_M\otimes1_N = 1_{M\otimes N}$. That is, can't we simply use the identification $(1_M\otimes\rho)\circ(1_M\otimes\phi) = 1_M\otimes 1_{N'} = 1_{M\otimes N'}$ to prove that $1_M\otimes\rho$ is a splitting $R$-homomorphism for the induced sequence whenever $\rho$ is for the first? Or am I missing something?

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  • $\begingroup$ What is $\rho$ ? $\endgroup$ – darij grinberg Aug 5 at 7:39
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    $\begingroup$ That said, you seem to be having the right idea: Rewrite the concept of a split exact sequence purely as a bunch of identities between morphisms (in your case, $\psi \circ \phi = 0$ and $\eta \circ \phi = \operatorname{id}$, where $\eta$ is a left inverse of $\phi$), and observe that functors (such as $M \otimes -$) must preserve such identities. $\endgroup$ – darij grinberg Aug 5 at 7:40
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    $\begingroup$ If $\rho\circ\phi=1$ then $(1\otimes\rho)\circ(1\otimes\phi)=1\otimes1=1$, so I think you are right. $\endgroup$ – drhab Aug 5 at 7:41
  • $\begingroup$ @darijgrinberg $\rho$ is a splitting $R$-homomorphism for $\phi$, that is, such an $R$-homomorphism for which we have $\rho\circ\phi = 1_{A'}$. $\endgroup$ – Jxt921 Aug 5 at 7:45
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    $\begingroup$ Oh, your sentence was too long for my current wakefulness :) Then, you're right. $\endgroup$ – darij grinberg Aug 5 at 7:46
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You are right, and the property is actually more general than that (if you know about functors, this works for any additive functor, not just $M\otimes-$, and the proof is the one you suggest)

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