$f \colon G_1 \to G_2$ group homomorphism, then $G_1 / N_1 \cong G_2 / N_2$ I'm stuck on a question regarding group theory, related to the isomorphism theorems. I am given a group homomorphism $f \colon G_1 \to G_2$, and $N_2$ a normal subgroup of $G_2$. I have proven that the pre-image $N_1 := f^{-1}(N_2)$ is a normal subgroup of $G_1$. Now, if we assume $f$ is surjective, I need to prove that $G_1 / N_1 \cong G_2 / N_2$.
I have already thought of using the first isomorphism theorem, but this has only lead to the conclusion that $G_1 / N_1 \cong G_2$, because $N_1$ is a normal subgroup and thus a kernel, and $f$ is surjective. Other than that, I have thought of using the canonical homomorphism $\varphi \colon G_2 \to G_2/N_2$. I suppose I could show the composition $\varphi \circ f$ is a bijection, but I'm not sure about how to prove this or whether this is the right direction for the proof. 
 A: First, observe that
\begin{align*}
\varphi : G_1 &\to G_2/N_2 \\
g_1 &\mapsto f(g_1)+N_2
\end{align*}
is a surjective group homomorphism.
Now, the identity of $G_2/N_2$ is $N_2$, and the inverse image of it is $N_1$.
So, the Kernel of the map $\varphi$ is $N_1$.
By $1$st isomorphism theorem, the result follows.
A: Let $\pi_i : G_i \rightarrow G_i / N_i$ be the two projection maps. Then if $\tilde{f} := \pi \circ f$, we have that $\tilde{f}$ is surjective, and that $ker\left(\tilde{f}\right) = N_1$. Then, by the isomorphism theorem, there is a map $\widehat{f} : G_1 / N_1 \rightarrow G_2 / N_2$ such that $\widehat{f} = \tilde{f} \circ \pi_1$. $\widehat{f}$ is injective by the theorem of isomorphism, and surjective because $\tilde{f}$ is.
A: Consider the commutative diagram:
\begin{alignat}{2}
G_1&\twoheadrightarrow&~& G_1/N_1\\
\downarrow~&&&\enspace\,\downarrow \\
G_2&\twoheadrightarrow&& G_2/N_2
\end{alignat}
The lower  way from $G_1$ to $G_2/N_2$  is surjective by hypothesis, so the upper way: $G_1 \to G_1/N_1 \to G_2/N_2$ is surjective, hence the second map is surjective.
