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I have been grappling with a lemma that doesn't seem like it should be too hard to solve, but has been giving me a run for my money, and I would really like some hints or general proof advice.

The statement of the lemma follows where $\cong$ means "isomorphic to":

Suppose $V$ and $V'$ are finite dimensional vector spaces with $U$ as a subspace of $V$ and $U'$ as a subspace of $V'$. If $V\cong V'$ and $U\cong U'$ then $$V/U\cong V'/U'$$

So at the outset, this seems very intuitive, even in an infinite dimensional case, but I was able to find a counter-example in my book, so I'm not too hung-up on that. But proving it has proven to be difficult, and I am now seeking help.

My proof, so far, follows:

Let $V$ and $V'$ be vector spaces over a field $\mathbb{F}$ such that $V\cong V'$ and let $U\subseteq V$ and $U'\subseteq V'$ be subspaces of $V$ and $V'$ respectively such that $U\cong U'$. I claim that since $V\cong V'$ and $U\cong U'$, then if we let $I_U$ be an isomorphism between $U$ and $U'$, then there is some isomprphism $I_V:V\to V'$ such that its restriction to $U$ follows $I_V|_U = I_U$ (aside: This is the part that I'm having trouble proving.)

If That is true, then consider the homomorphism $$ T : V\to V'/U'$$ $$ v\mapsto \{I_V(v)\} + U'$$ For the purpose of notation, define $[x]_{U'} := \{x\} + U'$ It's kernel is $U$, indeed if we let $x\in U$, then $$T(x) = [I_V(x)]_{U'} = [I_U(x)]_{U'} = [0]_{U'}\quad \text{since}\quad I_U(x) - 0\in U'$$ So $U\subseteq \ker(T)$. Now let $y\in\ker(T)$, so then $[I_V(y)]_{U'} = T(y) = [0]_{U'}$. This means that $I_V(y)\in U'$ now since $I_U$ is an isomorphism, there is some $u_0\in U$ such that $I_U(u_0) = I_V(y)$, but since $I_V|_U = I_U$, and $u_0\in U$, then $I_U(u_0) = I_V(u_0)$. So really, $I_V(u_0) = I_V(y)$, and since $I_V$ is an isomorphism, it's injective, so $y = u_0\in U$ so $\ker(T)\subseteq U$ so then $\ker(T) = U$.

I will now show that $T$ is surjective. Let $A\in V'/U'$, so then there is some $a\in V'$ such that $A = [a]_{U'}$. So then $I^{-1}_V(a)\in V$, now what is $T(I^{-1}_V(a))$? Well, it's the following: $$T(I^{-1}_V(a)) = [I_V(I^{-1}_V(a))]_{U'} = [a]_{U'} \in V'/U'$$ So then we can conclude from an isomorphism theorem, that: $$ V'/U'\cong V/\ker(T) = V/U$$

So far, nothing assumes that $V$ is finite dimensional, so I can surmise that I have to use it's finite dimensionality in showing that there is and isomorphism from $V$ to $V'$ such that it's restriction to $U$ is also an isomorphism, and this is where I am stuck. I only have fleeting thoughts as to what I could attempt.

So far I thought that since $V$ is a vector space and $U\subseteq V$, then there is some subset of the basis for $V$ that is a basis for $U$, call it $B_U$. Now lets define a homomorphism $I\in\mathcal{L}(V, V')$ Such that $$ I(b) \begin{cases} I_U(b) & b\in B_U \\ \text{something else} &b\in B\setminus B_U \end{cases} $$

So for what I was going to put into "someting else", I was thinking that since $B$ is finite, and $B_U$ is finite as well, then we could just have $I(b)$ for $b\in B\setminus B_U$ map to some vector that is not in the span of $I_V(B_U)$, but then at this point, I thought that maybe I'm just missing something simple and decided to make this post.

I hope that this hasn't been asked before, I tried searching but was unable to find a question about this.

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  • $\begingroup$ It's not true for infinite-dimensional vector spaces. $\endgroup$ – Lord Shark the Unknown May 21 '18 at 6:03
  • $\begingroup$ I did find a counterexample, and I'm not too worried that I intuited that it should work in an infinite dimension space. Now I just want some advice or a hint for how to finish the proof, or if I should abandon my current proof and try a different method. $\endgroup$ – hLance May 21 '18 at 6:19
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    $\begingroup$ I suggest counting dimensions. $\endgroup$ – Lord Shark the Unknown May 21 '18 at 6:28
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    $\begingroup$ In fact it's arguably because it fails for infinite dimensional spaces that you should look for a proof that uses finite bases... which just reduces essentially to counting the dimensions. $\endgroup$ – T_M May 21 '18 at 6:35
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Hint:

Since $U' \cong U$, $\dim U' = \dim U$, and $\dim V/U = \dim V' / U'$. Any what do we know about two vector spaces over the same field having the same dimension ?

Edit:

You can explicitly write an isomorphism between those spaces by mapping a basis of one to another, which, basically, boils down to proving the more general case the above lemma.

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  • $\begingroup$ In the book that I'm reading, they introduce the lemma in my original post before the theorem that vector spaces with the same dimension are isomorphic, so I'm trying to prove the above lemma without invoking that theorem. $\endgroup$ – hLance May 21 '18 at 6:43
  • $\begingroup$ @hLance Then first prove the above lemma yourself, and then evoke that lemma ? :) $\endgroup$ – onurcanbektas May 22 '18 at 5:33
  • $\begingroup$ @hLance The author probably expect you to write explicitly an isomorphism between those spaces by mapping the basis of one to another, which is the same thing as first proving the more general case above lemma because the logic in both proofs are the same. $\endgroup$ – onurcanbektas May 22 '18 at 5:36
  • $\begingroup$ Yeah finding a bijection between some bases of V and V' is a good idea. It's just that finding a bijection between the 2 spaces whose restriction to U is also a bijection between U and U' is proving to be difficult. $\endgroup$ – hLance May 23 '18 at 23:18

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