Proof of first isomorphism theorem of group 
(Proof of (first) Isomorphism Theorem) Let $f : G \rightarrow H$ be a surjective group homomorphism.
  Let $K = \operatorname{ker} f$. Then the map $f' : G/K \rightarrow H$ by $f'(gK) = f(g)$
  is well-defined and is an isomorphism. Proof: If $g'K = gK$, then $g' = gk$ with $k ∈ K$, and $f(g') = f(gk) = f(g) f(k) = f(g) e = f(g)$ so the map $f'$ is well-defined. It is surjective because $f$ is. For injectivity, if $f'(gK) = f'(g'K)$, then
  $f(g) = f(g')$, and $e_{H} = f(g)^{-1} · f(g') = f(g^{−1}) · f(g) = f(g^{−1}g')$
  Thus, $g^{−1}g' ∈ K$, so $g' ∈ gK$, and $g'K = gK$.
  In summary, the normal subgroups of a group are exactly the kernels of surjective homomorphisms.

First of all, I am not getting how $f(g)^{-1} \cdot f(g')$ leads to $f(g^{-1}) \cdot f(g)$.
Secondly, I am unsure how the later part of the proof establishes injectivity.
And lastly, at http://en.wikipedia.org/wiki/Isomorphism_theorem#Discussion, there is a diagram that shows the relationship, and can anyone explain this diagram with some insights from the above proof?
 A: For your first question, it is essentially assumed by hypothesis and the definition of $f'$ that $f(g')=f(g)$. But recall that for a homomorphism $f$, $f(g)^{-1}=f(g^{-1})$. This can be seen since $f(g^{-1})f(g)=f(g^{-1}g)=f(e_G)=e_H$, so $f(g^{-1})$ is indeed a  left inverse for $f(g)$, and the same argument shows it is a right inverse. So just substitute to conclude $f(g)^{-1}\cdot f(g')=f(g^{-1})\cdot f(g)$.
Secondly, writing $x=gK$ and $y=g'K$, the proof shows that if $f'(x)=f'(y)$, then $x=y$. This is the standard method to show a function is injective. Another way to look at it is the equivalent statement that if $x\neq y$, then $f'(x)\neq f'(y)$. So $f'$ always takes distinct inputs to distinct outputs, that is, $f'$ is injective.
A: I like to satate the next lemma to prove the first isomorphism theorem, if $f:G\longrightarrow H$ is a group morphism and $K\leq \ker f$ with $K$ normal subgroup of $G$, the map $\bar{f}:G/K\longrightarrow H$ given by $\bar{f}(gK)=f(g)$ is well defined and a group morphism and $\ker\bar{f}=\ker f/K$. 
To prove that is well defined, let $gK=hK$ then $h^{-1}g\in K\subseteq\ker f$, so $f(h^{-1}g)=e$, from here $f(g)=f(h)$.
