For a fixed Abelian group $G$, any map $f:A\to B$ induces a homomorphism $$f\otimes \operatorname{id}\colon A\otimes G \to B\otimes G.$$
Show that this defines a right exact functor, i.e. for any sequence $$A \to B \to C \to 0$$ the induced map $$A\otimes G \to B\otimes G \to C\otimes G \to 0$$ is also exact.
I am currently working through the solution, but I don't understand it. I have several questions. The solution says
Since the $0$ map induces the $0$ map on tensor products and $(f\otimes \operatorname{id})\circ (g\otimes\operatorname{id}) = (f \circ g) \otimes \operatorname{id}$, we see that any two successive maps still compose to the trivial map.
Ok, this makes sense but why do i need it in this case? And shouldn't it be $g\circ f$ rather than $f\circ g$ ?
This is where I have most trouble following:
Also $f\otimes \operatorname{id}$ is clearly surjective if $f$ is. Thus it sufices to show that the map $B \otimes G/\operatorname{im}(f\otimes \operatorname{id}) \to C \otimes G$ is an isomorphism.
Why is that? Why does it sufice to show that $B \otimes G/\operatorname{im}(f\otimes \operatorname{id}) \to C \otimes G$ is an isomorphism to prove exactness?
Suppose $f:A\to B$ is surjective. Then for $b\otimes g\in A\otimes G$, if $f(a)=b$, we have $(f\otimes 1)(a\otimes g)=f(a)\otimes g=b\otimes g$, so $f\otimes 1$ is surjective. That takes care of this part. Now we need the fact that $\mathrm{im}(f\otimes 1)= \mathrm{ker}(g\otimes 1).$
Yes you are right that it should be $g\circ f$ instead of $f\circ g$. You need that $(g\otimes 1)\circ (f\otimes 1)=0$ because this shows that $\mathrm{im}(f\otimes 1)\subset \mathrm{ker}(g\otimes 1)$. Now you need to prove the reverse inclusion in order to complete the proof.
Showing that the induced map $(B\otimes G)/\mathrm{im}(f\otimes G)\to C\otimes G$ is an isomorphism suffices. Indeed, since $g\otimes 1$ is surjective (because $g$ is), we have $C\otimes G\cong (B\otimes G)/\mathrm{ker}(g\otimes 1)$ by the first (?) isomorphism theorem. If $\mathrm{im}(f\otimes 1)$ were a strict subset of $\mathrm{ker}(g\otimes 1)$, then the induced map $$(B\otimes G)/\mathrm{im}(f\otimes G)\to C\otimes G\cong (B\otimes G)/\mathrm{ker}(g\otimes 1)$$ would be a quotient, actually the quotient by $\mathrm{ker}(g\otimes 1)/\mathrm{im}(f\otimes 1)$.