Surjective module homomorphism? $0$ module homomorphism?

Question

I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:

Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A \rightarrow B$ and $g:B\rightarrow C$ are both $R$-module morphisms.

1) $\forall \phi : C \rightarrow P$ morphism, if $\phi \circ g = 0 \Rightarrow \phi = 0$, for a morphism $\phi : C \rightarrow P$, does this imply that $g$ is surjective? Why?

2) If $\phi \circ g \circ f = 0$ $ \forall \phi : C \rightarrow P$ morphism does this mean that $g \circ f = 0$? Why?

Answer 1

For the first point consider for $\phi$ the quotient morphism $\pi:\ C\ \longrightarrow\ \operatorname{coker}g$.

For the second point consider for $\phi$ the identity morphism $\operatorname{id}:\ C\ \longrightarrow\ C$.

Answer 2

1) No, not necessarily. Here's a counterexample. Let $B=\mathbb{Z}$ and $C=P=\mathbb{Q}$ as $R=\mathbb{Z}$-modules. Further, let $g:B\to C$, i.e. $g:\mathbb{Z}\to\mathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $\varphi:C\to P$, i.e. for any $\varphi:\mathbb{Q}\to\mathbb{Q}$, $\varphi\circ g=0$ implies $\varphi=0$ (basically, because $\varphi\circ g=0$ implies $\varphi(1)=0$ implies $\varphi=0$). And yet, $g\neq0$.

2) No, not necessarily. Here's a counterexample. Let $A=B=C=\mathbb{Z}$ and $P=\mathbb{Z}_2$ as $R=\mathbb{Z}$-modules. Further, let $f:\mathbb{Z}\to\mathbb{Z}$ be the identity map $f(n)=n$ and $g:\mathbb{Z}\to\mathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $\varphi:\mathbb{Z}\to\mathbb{Z}_2$. we have $\varphi\circ g\circ f=0$, even though $g\circ f\neq0$.