Basic question about the image and kernel of tensor products I have a very basic question about the interactions of the image/kernel of module maps with tensor products.
For a ring $A$, $A$-modules $M, M'$ and $N$, and a map of $A$-modules $f: M\mapsto M'$, do the following hold?
$$\operatorname{Im}(f \otimes_A Id_N) = \operatorname{Im}(f) \otimes_A N $$
$$\operatorname{Ker}(f) \otimes_A N \subseteq  \operatorname{Ker}(f \otimes_A Id_N)$$
 A: Yes. For your first question, we first show the inclusion $\subseteq$. Let $x\in M\otimes_A N$ be arbitrary; then there are $m_i\in M$ and $n_i\in N$ such that $x=\sum_{i=1}^km_i\otimes n_i$. Then we have $$(f\otimes 1_N)(x)=\sum_{i=1}^k(f\otimes 1_N)(m_i\otimes n_i)=\sum_{i=1}^kf(m_i)\otimes n_i.$$ Each $f(m_i)\otimes n_i\in \operatorname{Im}{(f)}\otimes_A N$, so $(f\otimes 1_N)(x)$ is also, as desired.
Now, to see the inclusion $\supseteq$, since $\operatorname{Im}(f)\otimes_A N$ is generated as an abelian group by simple tensors of the form $f(m)\otimes n$, it suffices to show that each $f(m)\otimes n\in\operatorname{Im}(f\otimes 1_N)$. But this is immediate, since $$f(m)\otimes n=(f\otimes 1_N)(m\otimes n).$$

For your second question, since $\operatorname{Ker}(f)\otimes_A N$ is generated as an abelian group by simple tensors of the form $m\otimes n$, where $m\in\operatorname{Ker}(f)$, it suffices to show that all such elements are in the kernel of $f\otimes 1_N$. But this holds immediately: $$(f\otimes 1_N)(m\otimes n)=f(m)\otimes n=0\otimes n=0\otimes 0,$$ as desired.
Note, the reverse inclusion does not hold in general. Indeed, let $A$, $M$, and $M'$ all equal $\mathbb{Z}$, take $N=\mathbb{Z}\big/2\mathbb{Z}$, and consider the map $f:\mathbb{Z}\rightarrow\mathbb{Z}$ given by $x\mapsto 2x$. Now, the induced map $$f\otimes 1:\mathbb{Z}\otimes(\mathbb{Z}\big/2)\rightarrow\mathbb{Z}\otimes(\mathbb{Z}\big/2)$$ sends $x\otimes\overline{y}$ to $2x\otimes\overline{y}=x\otimes2\overline{y}=x\otimes \overline{0}=0\otimes \overline{0}$, and so the kernel of $f\otimes 1$ is the whole space. On the other hand, $\operatorname{Ker}(f)=\{0\}$, so $\operatorname{Ker}(f)\otimes(\mathbb{Z}\big/2)$ is the trivial submodule.
