Why do we have the following implication if $\phi$ is injective If $\phi: G \rightarrow H$ is a homorphism, and if $\phi$ is injective, why do we have the following:
$\phi(g) = e_h \implies g=e_g$
 A: Suppose $\phi(g)=e_{H}$. Since $\phi$ is a homomorphism, we know that $\phi(e_{G})=e_{H}$. Thus, by injectivity, $\phi(e_{G})=\phi(g)\implies e_{G}=g$.
A: $\phi(g^2)=\phi(g)^2=e_H^2=e_H=\phi(g)$ hence $g^2=g$ by injectivity. 
And $g^2=g\implies g=e_G$.

Sidenote:
$\phi:G\to H$ is a grouphomomorphism if $\phi(ab)=\phi(a)\phi(b)$, so actually it is not demanded that $\phi(e_G)=\phi(e_H)$ as you would expect. The fact that $\phi$ respects multiplication implies that $\phi$ respects idempotents (i.e. if $g^2=g$ then also $\phi(g)^2=\phi(g^2)=\phi(g)$). Next to that it can be shown that every group has exactly one idempotent, which is its neutral element. So automatically the idempotent $e_G$ is sent to the (unique) idempotent $e_H$. This makes demanding $\phi(e_G)=\phi(e_H)$ redundant, and in that knowledge it is possible to give a shorter answer to your question as done by @ervx.
A: Recall that a homomorphism maps the identity to the identity.  Thus, if $\phi: G \rightarrow H$ is a homomorphism, then $\phi(e_G)=e_H$.  Thus, there is already one element in $G$ which is mapped to the identity in $H$.  If $\phi$ is injective, then (by definition of 1-1) there cannot be any other element that is also mapped to the identity $e_H$. Thus, $\phi(g)=e_H$ implies $g=e_G$. 
