Given a vector space $V$ and its subspace $W \subseteq V$. I know that there is a bijection between the set of all subspaces of $V$ that contain $W$ and the set of all subspaces of $V /W$, induced by the projection homomorphism $\pi(v) = v + W$ where $v \in V$. I have no problem proving that it is an injection, but I am having difficulty proving that the map is surjective.
Let $M \subseteq V/W$ be given. Then $\pi^{-1}(M) = \{v \in V | \pi(v) \in M \} \subseteq V$. So I know that $\pi^{-1}(M)$ is a subspace of $V$, but how do I know that this subspace must contain $W$? Does a subspace of $V/W$ always contain $W$? Also, is showing this enough to conclude that the map is subjective? Thanks a lot in advance!
EDIT: Thanks to the comment by @Vercassivelaunos, I realized that I was confusing the notion of quotient space with the vector space itself. Clearly, any subspace of $V/W$ contains the zero vector, whose preimage under projection is $W$. But I am still wondering whether this is sufficient to show that the map is surjective. Can someone comment on this? Thanks a lot in advance!
Best Answer
To say that $\pi^{-1}(M)$ contains $W$ is to say that $\pi(W) \subseteq M$. So take a generic element $w \in W$, and we want to see why $\pi(w) \in M$.
But as you said in the comments, the subspace $M \subseteq V/W$ must contain the zero vector $0_{V/W} = 0+W = W$. And so $\pi(w) = w + W = W \in M$ as we wanted.