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?
Best Answer
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)