Suppose we have two short exact sequences:
$$0 \to M' \mathrel{\overset{f}{\to}} M \mathrel{\overset{g}{\to}} M'' \to 0 $$ in Mod-R
$$0 \to N' \mathrel{\overset{h}{\to}} N \mathrel{\overset{k}{\to}} N'' \to 0 $$
in R-Mod. I'm trying to prove that:$$ ker(g \otimes k) = Im(f \otimes 1_{N}) + Im(1_{M} \otimes h) $$
I easily prove that: $ Im(f \otimes 1_{N}) + Im(1_{M} \otimes h) \subseteq Ker(g \otimes h)$
but I'm stuck with the other contention.
I have been trying with diagram chasing, using the right exactness of the tensor functor but I'm getting nowhere. I appreciate your help.
First note that if either $g$ or $k$ is an isomorphism, then the proposition is just the right exactness of tensor functor.
Now rewrite $g\otimes k$ as $(g\otimes N'')(M\otimes k)$, where we rewrite $1_M$ as $M$ for abbreviation, or note that $-\otimes N''$ is a functor, therefore applies to a morphism $g$. By right exactness of tensor functor, we have $\ker(g\otimes k)=(M\otimes k)^{-1}(f\otimes N'')(M'\otimes N'')$
Now it's a matter of diagram chasing. $$ \require{AMScd} \begin{CD} M'\otimes N'@>>>M'\otimes N@>>>M'\otimes N''@>>>0\\ @VVV@VVV@VVV\\ M\otimes N''@>>>M\otimes N@>>>M\otimes N''@>>>0 \end{CD} $$ where horizontal lines are exact.
I won't spell out the details, but mention that the fact that $f$ and $h$ are injective isn't used. Formally, given the following diagram $$ \require{AMScd} \begin{CD} A'@>>>A@>>>A''@>>>0\\ @VVV@VVV@VVV\\ B'@>>>B@>>>B''@>>>0 \end{CD} $$ with exact rows and $a''\in A'',b\in B''$ such that their images in $B''$ are the same, then there exist $a\in A$ and $b'\in B$ such that the sum of their images in $B$ equals to $B$.
It's a special case of acyclic assembly lemma, which is a simplest case of spectral sequence.