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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!

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    $\begingroup$ $M$ contains the zero vector, whose preimage under the canonical projection is precisely $W$. By the way, no subspace of $V/W$ contains $W$, since the quotient is an entirely different vector space from $V$, not a subspace of $V$. $\endgroup$ Jan 19, 2022 at 14:17
  • $\begingroup$ I see! I think I confused the quotient space and vector space itself (I somehow was thinking the preimage of W is W). Is it enough to conclude that the map is surjective though? $\endgroup$ Jan 19, 2022 at 14:40
  • $\begingroup$ @InsultedByMathematics the projection map is surjective because its range is $W$ and it stabilizes $W$. The quotient space will be isomorphic to the orthogonal compliment of $W$. You can sort of thinking of it as removing the $W$ part and being left with the orthogonal compliment. I recommend investigating projections of $\mathbb{R}^3$ onto a line and onto a plane. See if you can figure out what the preimage of those projections must be, and how the orthogonal compliment plays a role in the quotient group. $\endgroup$ Jan 19, 2022 at 15:03
  • $\begingroup$ I see! Thank you very much for your insight! But I don't quite understand the concept of stable maps. Is what I have here enough for showing that the projection is surjective? $\endgroup$ Jan 19, 2022 at 15:12
  • $\begingroup$ @Vercassivelaunos I just realized one thing: isn't the zero vector in $V/W$ $0 + W$ = $W$? $V/W$ is the set of cosets of $W$ right? I think that's why I got confused in the first place. $\endgroup$ Jan 19, 2022 at 21:26

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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.

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  • $\begingroup$ Thank you very much for your insight! Is this sufficient to prove that the map is surjective though? $\endgroup$ Jan 20, 2022 at 2:56
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    $\begingroup$ Yes: you have taken a generic subspace $M \subseteq V/W$ and shown that there always exists a preimage $\pi^{-1}(M)$ belonging to the set of subspaces of $V$ containing $W$. Thus the map $\{\text{subspaces of } V \text{ containing } W\} \to \{\text{subspaces of } V/W\}$ is surjective. $\endgroup$ Jan 20, 2022 at 16:14
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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

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

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  • $\begingroup$ Thank you very much for the general result! Indeed, I learned the quotient space as the set of cosets in my linear algebra class. But just out of curiosity, does what I have shown imply that the projection is surjective? $\endgroup$ Jan 20, 2022 at 8:24
  • $\begingroup$ @InsultedByMathematics The projection is surjective, because $v+W=\pi(v)$. $\endgroup$
    – egreg
    Jan 20, 2022 at 10:30

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