# Verifying a bijection

Let $V$ be vector space over $\mathbb{F}$, and $W\subseteq V$ a subspace. Let $p:V\rightarrow V/W$ be the canonical projection. Let $X$ be the set of all subspaces containing $W$ and $Y$ be the sets of all subspaces of $V/W$. I want to show that $p$ induces a bijection in the following way:

$L\in X$ is mapped to $p(L)=\{p(v) \mid v\in L\}$

$M \in Y$ is mapped to $p^{-1}(M)=\{v\in V \mid p(v) \in M\}$.

I feel as though I'm just having trouble with the set theory. Namely, I think I'm on the right track by just showing that each one of these inverts the other, but I'm having trouble sifting through the notation...

Help with what I just mentioned, and getting some intuition for what this map is telling me would be most appreciated!

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If you already learned some group theory this is the famous "Correspondence theorem". Even if you haven't learned group theory you can search for it and try to fix it to vector spaces. – DonAntonio Sep 28 '12 at 5:44

Let $L \in X$. We claim that $L = p^{-1}p(L)$. Clearly $L \subset p^{-1}p(L)$. Let $x \in p^{-1}p(L)$. Then $p(x) \in p(L)$. Then $p(x) = p(y)$, where $y \in L$. Hence $x - y \in p^{-1}(0) = W$. Hence $x \in L + W = L$. Hence $p^{-1}p(L) \subset L$.

Let $M \in Y$. We claim that $M = pp^{-1}(M)$. Clearly $pp^{-1}(M) \subset M$. Let $y \in M$. Since $p$ is surjective, there exists $x \in V$ such that $p(x) = y$. Then $x \in p^{-1}(M)$. Hence $y \in pp^{-1}(M)$. Hence $M \subset pp^{-1}(M)$.

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In $V$ the elements of $V/W$ are subsets of $V$; in $V/W$ they’re points. Thus, the quotient is really just formed by chopping $V$ up into (nice) chunks and wadding each chunk up to make a single point. The map $p$ then sends each point of $V$ to the chunk containing it. Here’s a quick and dirty picture that may help you to visualize more easily what’s going on:

If $V$ is on the blackboard, $p$ is basically just squashing everything down into the chalk tray and collecting each coset of $W$ into a single lump of chalk dust.

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That helped a lot with the intuition! Thanks so much, sir. – AsinglePANCAKE Sep 28 '12 at 6:15
@AsinglePANCAKE: Glad to help! – Brian M. Scott Sep 28 '12 at 6:15