Could I prove this coset theorem using just homomorphism like following? In Dummit's Abstract Algebra, there is this Proposition 3.4 in page 80:

If there is some homomorphism from G to H as following:

Could I prove it by using only that homomorphism as following?
That homomorphism would bring $u$, $v$, $n$, and $m$ to exactly only one element in the codomain H: $U$, $V$, $N$, and $N$ respectively (otherwise it's not a function). So $UN = VN$ is only when $U = V$. As homomorphism only brings any values from domain to exactly only one value in codomain, uN = vN is only when u and v are in the same fiber of homomorphism. I said in the same fiber because also note that in my proof the set of elements of fiber above N is even not necessarily a subgroup of $G$. Where am I wrong?
 A: I think you want to define a homomorphism $\phi: G \to \{\text{left cosets of }N \text{ in }G\}$, by saying that given an element $g \in G$ you can write $g= un$, for some $u \in G \setminus N$ and $n \in N$ and then define $\phi(g)=u$. 
This works whenever $N$ is normal, it is the standard surjective map $\pi: G \to G/N$. However, it never works when $N$ is not normal (as celtschk pointed out in the comments, you can actually prove that $N$ is normal when you have such a homomorphism, since $N = \phi^{-1}(1)$).
The reason it does not work is because multiplication fails: the left cosets partition is not a group (at least not with a structure inherited from $G$). In fact, given $g=u_g n_g$, $h= u_h n_h$ then it is not true in general that $gh$ is in the $u_gu_h$ coset!
Write
$$u_g n_g u_h n_h = u_g u_h u_h^{-1} n_g u_h n_h = u_g u_h (u_h^{-1} n_g u_h) n_h$$
Now if this element were in the $u_g u_h$ coset you'd have $u_h^{-1}n_gu_h \in N$, which holds when $N$ is normal, but not in general. 
In a nutshell, the left cosets partition is not well-behaved with respect to the group operation unless $N$ is normal, which is why you only have a group structure (and also a homomorphism, which is not hard to prove once you know the above) when $N$ is normal.
A: Let's write $u\sim v$ iff $u\in vN$.
There indeed exists a canonical function $\varphi:G\to G/\sim$ with fibers exactly the left cosets, but


*

*In order to establish the quotient set $G/\sim$ we should already know that $\sim$ is an equivalence relation, which is basically the claim of this proposition.

*We will get a group structure on $G/\sim$ iff $N$ is a normal subgroup, which means that the left and right cosets coincide. (Observe e.g. that taking elementwise inverses, we obtain $(gN)^{-1}=N^{-1}g^{-1}=Ng^{-1}$.)
In this case, $\varphi$ will be indeed a homomorphism. 
But this is not required for proving that $\sim$ is an equivalence relation.

Note also, that for any subset $N$ of $G$, we have the following facts about the relation $\sim$ defined above:


*

*$\sim$ is reflexive iff $1\in N$ (where $1$ denotes the identity element of $G$).

*$\sim$ is symmetric iff $N$ is closed under inverses ($x\in N\implies x^{-1}\in N$).

*$\sim$ is transitive iff $N$ is closed under multiplication ($x, y\in N\implies xy\in N$).

