$m \subset R$ is maximal $\iff$ $0 \subset R / m$ is maximal Let $R$ be a ring and $m$ be a maximal ideal. The goal is to show $m \subset R$ is maximal $\iff$ $0 \subset R / m$ is maximal.
The proof is given as: Let $\pi: R \to R/m$, $r \mapsto r +m$.
Then is
$$\Rightarrow \; \{I \subset R \, \vert\;m \subset I \;\text{ideal}\, \} \to \{J \subset R/m \,\vert \, J \;\text{ideal}\;\}, \; I \mapsto \pi(I)$$
a bijective map. Hence we have that $m \subset R$ maximal $\iff$ $0=m/m \subset R/m$.
The last part is quite fast. I don't understand why this map should be bijective, and how can we conclude the proof with this?
Many thanks for some help!
 A: If $I\subseteq R$ is an ideal which contains $m$ then $\pi(I)$ is an ideal of $R/m$. We'll show this correspondence is bijective.
Surjective: Suppose $J\subseteq R/m$ is an ideal. It is easy to check that $\pi^{-1}(J)$ is an ideal of $R$ which contains $m$. Since $\pi$ is a surjective map, the equality $J=\pi(\pi^{-1}(J))$ holds.
Injective: Suppose $I\subseteq R$ is an ideal which contains $m$. We'll show that $\pi^{-1}(\pi(I))=I$. Obviously, $I\subseteq\pi^{-1}(\pi(I))$, this is immediate from the definitions. Conversely, if $x\in\pi^{-1}(\pi(I))$ then $\pi(x)\in\pi(I)$, and so there is some $i\in I$ such that $x+m=\pi(x)=\pi(i)=i+m$. Then $x-i\in m\subseteq I$, and so $x=i+(x-i)\in I$, as a sum of elements from $I$.
It follows that if $I_1, I_2\subseteq R$ are ideals which contains $m$ such that $\pi(I_1)=\pi(I_2)$ then:
$I_1=\pi^{-1}(\pi(I_1))=\pi^{-1}(\pi(I_2))=I_2$
And so the correspondence is also injective.
So we have shown there is a bijection between the ideals of $R/m$ and the ideals of $R$ which contains $m$. In particular, if $m$ is a maximal ideal then there are only $2$ ideals of $R$ which contain it, namely $m$ and $R$. In that case, the only ideals of $R/m$ are the trivial ideals, and so the zero ideal of this ring is maximal. The converse is similar.
