TITLE: Proof of first isomorphism theorem of group
QUESTION [7 upvotes]: (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?

REPLY [2 votes]: 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 $f(g)=f(h)$.