Proving that $\tilde{T}$ is 1-1 & onto For $v \in V$, we use $[v]$ to denote the equivalence class containing $v$. Let $T: V \to W$ be a linear map. Defining the linear map $\tilde{T}: V/\ker(T) \rightarrow \operatorname{im}(T)$ with the formula $\tilde{T}([v]) = T(v)$, we want to show that $\tilde{T}$ is 1-1 & onto.
If $v \sim w \in \ker(T)$, then we know $T(v - w) = 0$, which also shows that $T(v) = T(w)$. I'm not sure how to proceed from here. Any help would be very appreciated. I'm pretty sure this has bearing on the first isomorphism theorem, but again I am not sure.
 A: What you're trying to prove is actually the first isomorphism theorem! You have correctly shown that $\tilde{T}$ is a well-defined mapping. Now we need to show that $\tilde{T}$ is also a linear map that is both injective (1-1) and surjective (onto). The linearity of $\tilde{T}$ is actually pretty easy to verify:
$$\tilde{T}([v]+[w]) = \tilde{T}([v+w]) = T(v+w)=T(v)+T(w) = \tilde{T}([v]) + \tilde{T}([w])$$
and
$$\tilde{T}(a\cdot[v]) = \tilde{T}([av]) = T(av) = a\cdot T(v) = a \cdot \tilde{T}([v])$$
For injectivity, let $[v] \in \ker \tilde{T}$. Then $\tilde{T}([v]) = T(v) = 0$ and therefore $v \in \ker T$ which implies that $[v] = [0]$. Since $\ker \tilde{T} = \{[0]\}$ we conclude that $\tilde{T}$ is injective.
For surjectivity, notice that for every $w \in \text{im}T$ we have some $v \in V$ such that $T(v)=w$. By definition, $\tilde{T}([v]) = T(v) = w$ which of course gives us the surjectivity of $\tilde{T}$.
We have just proven that there exists an isomorphism $V/\ker T \simeq \text{im}T$ which is exactly what the first isomorphism theorem states.
