Let $f:G \rightarrow H$ denote a surjective homomorphisms of groups. Then $\bar{f}: G/ker(f) \rightarrow H$ is an isomorphism
Proof: We have to first show that the map is well-defined. In particular, suppose $gK = g'K$ where $K = ker(f)$. Then we have to show that $\bar{f}(g) = \bar{f}(g')$. In particular, we have to show that $f(g) = f(g')$. But, $gK = g'K$ is equivalent to $g'^{-1}g \in K$. It follows that $e = f(g'^{-1}g)=f(g')^{-1}f(g)$ and therefore $f(g') = f(g)$. Next, one can easily check that $\bar{f}$ is a homomorphism. Finally, we have to check that it is injective and surjective. Let $h \in H$. Then, since $f$ is surjective, there is a $g \in G$ such that $f(g) = h$. Then, by definition, $\bar{f}(gK) = h$, and therefore $\bar{f}$ is surjective. For injectivity, it is enough to show that $ker(\bar{f}) = eK$. Let $gK \in ker(\bar{f})$. Then $e = \bar{f}(gK) = f(g)$. It follows that $g \in K$, but this is equivalent to $gK = eK$. $\blacksquare$
Questions:
Why do we need to show $f(g) = f(g')$? Isn't this already a well defined function?
Why does it follow by definition of $f$ being surjective that $\bar{f}$ is surjective?
Why is it enough to show injectivity by showing that $ker(\bar{f}) = eK$. This only shows that there is only one value, namely the identity, that maps to the identity. I do not see how this generalizes to all the elements.
How is $e = \bar{f}(gK) = f(g)$ equivalent to $gK = eK$?
$1.$ How can we be sure it is well defined? The function is defined on a set of equivalence classes. For the left coset $gK$ we have $\bar{f}(gK)=f(g)$. So $f$ acts on the coset $gK$ and returns something that depends on the element $g$. Now what if we have $gK=g'K$ and we find out that $f(g)\ne f(g')$? It would mean that $\bar{f}$ would return two different values for the same coset, which would mean it is not a well defined function. So we must check that if $gK=g'K$ then $\bar{f}(gK)=\bar{f}(g'K)$.
$2.$ Let $h\in H$. Since $f$ is surjective there is some $g\in G$ such that $f(g)=h$. Then $\bar{f}(gK)=f(g)=h$. So $\bar{f}$ is surjective.
$3.$ If $\varphi:G\to H$ is any homomorphism of groups then it is injective if and only if the kernel is trivial, this is a well known theorem. Suppose $\varphi(g_1)=\varphi(g_2)$. Then $e_H=\varphi(g_1)(\varphi(g_2))^{-1}=\varphi(g_1g_2^{-1})$, and hence $g_1g_2^{-1}\in Ker(\varphi)$. Since we supposed the kernel is trivial this implies $g_1g_2^{-1}=e_G$, and hence $g_1=g_2$. So $\varphi$ is indeed injective.
$4.$ Since $f(g)=e$ it follows that $g\in K$, and hence $e^{-1}g\in K$. This by definition means that $gK=eK$.