Tensor product and exact sequences

Solution 1:

Good question! I got really confused about this as well.

I thought of this "proof" of the fact $\mathrm{Im}(f\otimes 1_N)=\mathrm{Im}(f)\otimes_R N$:

$\subseteq$: We have $(f\otimes 1_N)(x\otimes a)=f(x)\otimes a\in \mathrm{Im}(f)\otimes_R N$ for $x\in M_1$ and $a\in N$.

$\supseteq$: We have $f(x)\otimes a=(f\otimes 1_A)(x\otimes a)\in \mathrm{Im}(f\otimes 1_N)$ for $x\in M_1$ and $a\in N$.

However, the problem here is that $\mathrm{Im}(f\otimes 1_N)$ and $\mathrm{Im}(f)\otimes_R N$ don't live in the same space! While $\mathrm{Im}(f\otimes 1_N)\subseteq M_2\otimes_R N$, the space $\mathrm{Im}(f)\otimes_R N$ doesn't live in $M_2\otimes_R N$, since $-\otimes_RN$ doesn't necessarily preserve the injection $\mathrm{Im}(f)\hookrightarrow M_2$!

What is true is that there is a $R$-bilinear map \begin{align*}\mathrm{Im}(f)\times N&\to \mathrm{Im}(f\otimes1_N)\\(f(x),a)&\mapsto f(x)\otimes a=(f\otimes 1_N)(x\otimes a),\end{align*} which induces a surjective $R$-homomorphism $\mathrm{Im}(f)\otimes_RN\to \mathrm{Im}(f\otimes 1_N)$.

I wonder whether the kernel of this map can be expressed in terms of the Tor functor...