Does $f\otimes \operatorname{Id} = \operatorname{Id}$ imply $f= \operatorname{Id}$? > Let $R$ be a commutative ring, and $X$ an $R$-module. If an $R$-endomorphism of $X$ satisfies $f\otimes \operatorname{Id}_X = \operatorname{Id}_{X\otimes X}$, is it true that $f=\operatorname{Id}_X$ ? It is true if $X$ is locally free, or monogenous, but I suspect it is false in general ; but I can't seem to come up with any convincing example. If by any chance it is actually true for module categories, is it also true for any symmetric monoidal category (seems even more unlikely, though) ?

Your guess is right, it's indeed not true in general: Any choice of an $R$-module $X$ such that $X\otimes_R X=0$ and $f\
eq\text{id}: X\to X$ gives a counterexample, for example you could take $R := {\mathbb Z}$, $X := {\mathbb Q}/{\mathbb Z}$ and $f = 2\cdot \text{id}$.