Normal subgroup involving quotient group 
Let $G$ be a group and $N$ be a normal subgroup of G.   Show that there
  is a bijection between $N\times G/N$ and $G$.

I think I can define a map by $f : N\times G/N\to G$ where $(n,gN) \mapsto ng$.
Then $f$ is surjective. But how can I show that $f$ is injective? Or is it possible to find its inverse function?

Let $G$ be a group, $N\triangleleft G$, and $G_{i}$, $G_{i+1}$ be two
  subgroups of $G$ such that $G_{i}\triangleleft G_{i+1}$.
  Then ($G_{i+1}\cap N$)/($G_{i}\cap N$) $\triangleleft$
$G_{i+1}$/$G_{i}$

I only know ($G_{i+1}\cap N$)/($G_{i}\cap N$) is well-defined. But I don't know what else I can do now.
 A: I will prove the first statement. The proof of the second one is not short (but it's standard) and it needs to be asked in a separate question because it's irrelevant to the first statement.
From basic definitions in group theory, we know that the projection map $\pi: G \to G/N$ is well-defined and it's surjective. By the axiom of choice, it must have at least one right inverse. Take a right inverse of $\pi$ and call it $\phi: G/N \to G$. Therefore, $\pi \circ \phi=\mathbb{1}_{G/N}$. Note that $\phi$ has the property that it is constant on each equivalence class because of the way we constructed it. Now, define $f: N \times G/N \to G$ by
$$f(n,gN)=n\phi(gN)$$
By construction, $g_1N=g_2N$ gives that $\phi(g_1N)=\phi(g_2N)$. So, it's seen that $f$ is well-defined.
$f$ is also surjective because for any $g\in G$, there exists some $n_0 \in N$ such that $g\phi(gN)^{-1}=n_0$ because they belong to the same equivalence class. Hence, $g=f(n_0,gN)$.
To check injectivity, note that $f(n_1,g_1N)=f(n_2,g_2N)$ gives that
$$n_2^{-1}n_1=\phi(g_2N)\phi(g_1N)^{-1}$$
$$N=\pi(n_2^{-1}n_1)=\pi(\phi(g_2N)\phi(g_1N)^{-1})=\pi(\phi(g_2N))\pi(\phi(g_1N))^{-1}=(g_2N)(g_1^{-1}N)=g_2g_1^{-1}N$$
Hence, $g_2g_1^{-1}\in N$ and they belong to the same equivalence class. In other words, $\phi(g_1N) = \phi(g_2N)$. Hence, $n_1=n_2$ and $f$ is injective too.
