# 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!

• $M$ 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 $V$. Jan 19, 2022 at 14:17
• 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? Jan 19, 2022 at 14:40
• @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. Jan 19, 2022 at 15:03
• 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? Jan 19, 2022 at 15:12
• @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. Jan 19, 2022 at 21:26

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.

• Thank you very much for your insight! Is this sufficient to prove that the map is surjective though? Jan 20, 2022 at 2:56
• 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. Jan 20, 2022 at 16:14

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 $$\ker\pi$$.

• 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? Jan 20, 2022 at 8:24
• @InsultedByMathematics The projection is surjective, because $v+W=\pi(v)$. Jan 20, 2022 at 10:30