Split exact sequence induced by tensor product: is my proof correct?

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?

• What is $\rho$ ? – darij grinberg Aug 5 at 7:39
• 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. – darij grinberg Aug 5 at 7:40
• If $\rho\circ\phi=1$ then $(1\otimes\rho)\circ(1\otimes\phi)=1\otimes1=1$, so I think you are right. – drhab Aug 5 at 7:41
• @darijgrinberg $\rho$ is a splitting $R$-homomorphism for $\phi$, that is, such an $R$-homomorphism for which we have $\rho\circ\phi = 1_{A'}$. – Jxt921 Aug 5 at 7:45
• Oh, your sentence was too long for my current wakefulness :) Then, you're right. – darij grinberg Aug 5 at 7:46

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)