How to show that when the linear transformation $\Phi =\psi \circ \phi : E \rightarrow G$ is injective, then $\phi$ is also injective? How to show that when the linear transformation $\Phi =\psi \circ \phi : E \rightarrow G$ 
(where $\phi : E \rightarrow F$ and $\psi : F \rightarrow G$) is (1) injective, then $\phi$ is injective and when (2) $\Phi$ is surjective, then $\psi$ is also surjective ?
I've tried to prove it by the definition of injection and surjection, but I'm not convinced myself by it.
For example, in the first case :
If $\Phi$ is injective, $\forall g \in G$, we have $0$ or $1$ $e \in E$ such as $\Phi(e)=g$.
$\Rightarrow \Phi(e) = \psi(\phi(e)) = \psi(f) =g$. 
So $\forall f \in F$, we also have $0$ or $1$ $e \in E$ such as $\phi(e)=f \in F$. 
$\Rightarrow \phi$ is injective.
Is this right or did I prove nothing at all ?
I think it is easier to convince myself/ to understand with surjectivity but still..
 A: Say $\phi (e_1)=\phi (e_2)$. Then $\psi \circ \phi (e_1)= \psi \circ \phi (e_2)$  and form injectivity of $\psi \circ \phi$ we have $e_1=e_2$. So $\phi $ is injective.
Now say $g\in G$. Then from surjectivity of $\psi \circ \phi$ we have $\exists e\in E$ such that $\psi \circ \phi (e)=g$  and  hence $\psi (\phi (e))=g$.  But $\phi(e)\in F$ and hence $\psi (\phi (e))=g$ shows that $\psi$ is surjective
A: What you know: $\Phi$ is injective. That is, $\forall x,y \in E$ $$\Phi(x)=\Phi(y) \implies x=y$$
What do you want to prove? That $\phi$ is injective. That is, $\forall x,y \in E$ $$\phi(x)=\phi(y) \implies x=y$$
Well, pick any pair $x,y$ such that $\phi(x)=\phi(y)$, and do $\psi$ to them. Since they are the same thing, $\psi$ must give the same output because it is a function. So $$\psi(\phi(x))=\psi(\phi(y)) \implies \Phi(x)=\Phi(y) \implies x=y$$
And you're done. The other proof is similar. Suppose $\Phi$ is surjective. Then $\forall g \in G$ there is an $e \in E$ with $\Phi(e)=g$. That is, $\psi(\phi(e))=g$, and $\phi(e) \in F$ so...
