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I'm trying to make sense of the following quotient: $(V/W)/(U/W)$. $V$ is a vector space over a field $\mathbb{F}$ and $W$ and $U$ are subspaces, specifically $W$ is a subspace of $U$. Naturally, the notation is suggestive that this double quotient is isomorphic to $V/U$, but I'm having some trouble proving this guess.

I construct the following square and try to make it commutative:

$$\begin{array}{rcl} V&\overset{p_1}\longrightarrow&V/W\\ \\ p_2\downarrow&&\downarrow p_3\\ \\ V/U&\underset{f}\longrightarrow&(V/W)/(U/W) \end{array}$$

And the bottom is my desired map, $f:V/U\rightarrow (V/W)/(U/W)$, where I define $$f(v+U)=p_3\big(p_1(v)\big)=p_3(v+W)\;.$$

Am I on the right track? I'm having some trouble showing that this is indeed a well-defined isomorphism...

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@BenMillwood Very true! I think I got them all... – AsinglePANCAKE Sep 23 '12 at 20:55
up vote 1 down vote accepted

Let $B_W$ be a basis for $W$. Extend it to a basis $B_U\supseteq B_W$ of $U$, and then extend that to a basis $B_V\supseteq B_U$ of $V$. Let $B_0=B_U\setminus B_W$ and $B_1=B_V\setminus B_W$; show that $\{b+W:b\in B_0\}$ is a basis for $U/W$, $\{b+W:b\in B_1\}$ is a basis for $V/W$, and $\{b+U:b\in B_1\setminus B_0\}$ is a basis for $V/U$.

Finally, show that $$B=\Big\{(b+W)+(U/W):b\in B_1\setminus B_0\Big\}$$

is a basis for $(V/W)/(U/W)$ and that the map $$\{b+U:b\in B_1\setminus B_0\}\to B:b+U\mapsto (b+W)+(U/W)$$ induces the desired isomorphism.

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So you have $W \subseteq U \subseteq V$, and you want to show that $(V/W)/(U/W) \cong V/U$.

For each $[v]_W \in V/W$ we have $[v]_W = \{ v + w : w \in W\},$ for each $[u]_W \in U/W$ we have $[u]_W = \{u + w : w \in W\}$ and for each $[v]_U \in V/U$ we have $[v]_U = \{v + u : u \in U\}.$

Consider the map $f : V/W \to V/U$ given by $[v]_W \mapsto [v]_U$. This map is a homomorphism since

$$f([v_1]_W+[v_2]_W) = f([v_1+v_2]_W) = [v_1+v_2]_U = [v_1]_U + [v_2]_U = f([v_1]_W)+f([v_2]_W) \, . $$

Moreover, $\ker(f) = U/W$ since $f([v]_W) = [u]_U$ if and only if $v \in U$.

A standard result from linear algebra tells us that if $g : A \to B$ is a linear homomorphism then $A/\ker(g) \cong \text{im}(A)$. It follows that $(V/W)/(U/W) \cong V/U$, as required.

Take a look at Wikipedia's article on the Isomorphism Theorem.

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It's not enirely obvious that $[v]_W \mapsto [v]_U$ is independent of the representative. – joriki Sep 23 '12 at 21:17

Yes, on the right track.

For $f$ being a well defined linear mapping, you need only to check that $0$ always goes to $0$, because of the presence of subtraction. (I.e., elements originally of $U$ goes to $U/W$).

For $f$ being injective, you need that the preimage of $0$ must be $0$ (here $W\subseteq U$ is needed).

Finally, $f$ is easily seen to be surjective.

Anyway, this argument is also valid for sets (or any algebraic structure). In this setting, $V$ is a set, and $U$ and $W$ are equivalence relations on it, such that $aWb$ implies $aUb$ for all $a,b\in V$.

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Mind elaborating a little more on why I only need to check 0 for well defined-ness? – AsinglePANCAKE Sep 23 '12 at 21:37
Suppose the same element in the quotient is represented by both $a$ and $b$, then use $f(a)-f(b)=f(a-b)$ to revert it back to the case of $0$. – Berci Sep 23 '12 at 23:17
I'm actually a bit confused by your last statement as well. My counterexample, which may be way wrong....Consider $V=\mathbb{R^3}$ $U$ the xy-plane and $W$ the x-axis. Surely aUb does not imply aWb. What confuses me is that it seems that is the implication I need for well-defined, and not the one you state. Again, I could be very messed up in my reasoning here... – AsinglePANCAKE Sep 24 '12 at 1:44

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