Proving the isomorphism between subspaces of a vector space and subspaces of quotient spaces 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!
 A: 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.
A: You can avoid dealing with quotient spaces by proving a more general result.
Theorem. If $f\colon V\to W$ is a linear map, then there is a bijection between the set of subspaces of $V$ containing $\ker f$ and the set of subspaces of $W$ cointained in $\operatorname{im}f$. The bijection is induced by the direct and inverse image with respect to $f$.
How to prove this? It's sufficient to prove that

*

*if $X$ is a subspace of $V$ and $X\supseteq\ker f$, then $f^\gets(f^\to(X))=X$;

*if $Y$ is a subspace of $W$ and $W\subseteq\operatorname{im}f$, then $f^\to(f^\gets(Y))=Y$.

Here $f^\to(A)=\{f(v):v\in A\}$ and $f^\gets(B)=\{v\in V:f(v)\in B\}$.
Note that $f^\gets(f^\to(X))\supseteq X$ with no assumption on $X$ and also $f^\to(f^\gets(Y))\subseteq Y$. Thus you just need to prove the reverse inclusions.
You shouldn't have problems in doing this: the special case when $f=\pi\colon V\to V/W$ follows. I think your issue is that you rely “too much” on the representation of elements in $V/W$ as cosets, so using $W$ in two “different” meanings: it is a subspace of $V$, but also the zero element in $V/W$. In the latter meaning it is $\pi(0)$, in the former meaning it is $\ker\pi$.
