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.
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.