Show an isomorphism between two quotient spaces

Question

We are given 3 vector spaces $U \subseteq W \subseteq V$

We are asked to prove that there is an isomorphism between $V/W$ and $(V/U)/(W/V)$

What we tried (and it is false, I would like to know why):

We basically said that an element of $V/W$ is $v+W$ and that an element of $(V/U)/(W/V)$ is $v+U+W+U = v+U+W = v+W$ and so we can define the identity transform, it is invertible, and so an isomorphism exists.

The teacher said it was incorrect. Would someone please tell me why and show me the right solution?

Answer 1

Hint: Use the formula $dim(A/B)=dim(A)-dim(B)$ if $B\subset A$.

About your solution you can't add cosets this way they are different objects.

Answer 2

Well, to begin with an element in$\;(V/U)/(W/U)\;$ is of the form $\;\left(v+U\right)+(W/U)\;$ (note that both sets of parenthese are there only for clearity purposes) and not what you wrote.

To prove that thing you can either use the good'ol first isomorphism theorem for groups (since vector spaces are, in particular, abelian groups), and just check that multiplication by scalar behaves well under this theorem, or directly: define

$$\phi: V/U\to V/W\;\;\;\text{by}\;\;\phi(v+U):=v+W .$$

1) Prove $\;\phi\;$ is well defined, i.e.: $\;v+U=v'+U\implies v+W=v'W\;$

2) Prove $\;\phi\;$ is a linear map of vector spaces

3) Prove $\;\ker\phi=W/U\;$

Answer 3

Both User52045's claims are right: proving they have the same dimensions is enough for finite dimensional vector spaces (over the same field). Also about you don't add cosets the way you do. Let's try to write an explicit isomorphism a little bit more correctly using cosets.

This isomorphism is just

$$ \phi : V / W \longrightarrow (V/U)/(W/U) \ , \qquad \phi (v + W) = (v+U)+ W/U \ . $$

First of all, you should check that such a map is well defined. That is, if you look carefully at its formula, it reads: you pick some representative $v$ of the class $v+W$, and send it to the class $(v+U)+ W/U$.

So, you have to ask: what if I pick some different $v'$ in the same class? That is, some $v' \in v+W \Longleftrightarrow v'+W = v +W$. But this would mean that $v'-v \in W$. Hence $(v'-v)+U \in W/U$. Therefore, $(v'-v)+U+W/U = 0$, inside $(V/U)/(W/U)$. Thus, by definition of addition in there, $(v'+U)+W/U = (v + U) +W/U$.

Secondly, you should prove that this $\phi$ is linear. For instance, that

$$ \phi (v + v') = \phi (v) + \phi (v') \ . $$

Which follows at once, again by definition of the addition in $(V/U)/(W/U)$. (Do it, and also that $\phi (\lambda v ) = \lambda \phi (v)$.)

Thirdly, let's see $\phi$ is injective: assume $\phi (v) = 0$. By definition of $\phi$, this means

$$ 0 = (v+U)+W/U \quad \Longleftrightarrow \quad v + U \in W/U \quad \Longleftrightarrow \quad v \in W \ , $$

which means $v+W = 0$ in $V/W$.

Finally, you should be able to check that $\phi $ is surjective. That is, if you pick any $(v+U)+W/U$ in $(V/U)/(W/U)$, you have to find some vector in $V/W$ that goes to your class. This should be easy, shouldn't it?