What is wrong with this proof of a function being injective?

Question

Okay we have an exact sequence

$0 \rightarrow A \xrightarrow[\text{}]{\text{f}} B \xrightarrow[\text{}]{\text{g}} C \rightarrow 0 $

where $A,B,C,$ and $D$ are $R$-modules with $R$ a commutative ring.

We are first asked to show that tensor product is right exact in the sense that

$A \otimes_{R}D \xrightarrow[\text{}]{f\otimes id} B \otimes_{R}D \xrightarrow[\text{}]{g\otimes id}C \otimes_{R} D \rightarrow 0$

and then give an example of a ring $R$ and $R$-modules $A,B,C,D$ such that the original sequence is exact but $f \otimes id$ is not injective. Note that $id$ denotes the identity function.

My (admittedly extremely trivial) argument is that $f$ is injective by exactness, hence has a left inverse which I will denote by $f^{-1}$ and then $f^{-1}\otimes id$ will serve as a left inverse for $f \otimes id$ so $f\otimes id$ also injective. Why does this not work? I'm assuming this is some sort of well defined issue due to equivalence classes.

Answer 1

While it is true that every injective map of sets has a left inverse, this is not true in the case of $R$-module homomorphisms $A\to B$.

For an example, take $R=A=B=\Bbb Z$, and the $\Bbb Z$-module (i.e. abelian group) homomorphism $\Bbb Z\to\Bbb Z$ sending $1\mapsto 2$. This is injective but does not have a left inverse.

Answer 2

Consider the canonical injection $i:pZ\rightarrow Z$, it does not have a left inverse $j:Z\rightarrow pZ$, if $j$ exists, you would have $j(p)=p=j(p.1)=p.j(1)$. The equation $px=p$ does not have a solution in $pZ$.