Background
This problem is taken from the book advanced calculus by h loomis and shlomo sternberg.
I have solved the first two problems:
2.9
A subspace $N$ of a vector space $V$ has finite codimension $n$ if the quotient space
$V/N$ is finite-dimensional, with dimension $n$. Show that a subspace $N$ has finite
codimension $n$ if and only if $N$ has a complementary subspace $M$ of dimension $n$.
(Move a basis for $V/N$ back into $V$.) Do not assume $V$ to be finite-dimensional.
and
2.10
Show that if $N_1$ and $N_2$ are subspaces of a vector space $V$ with finite codimensions,
then $N=N_1\cap N_2$ has finite codimension and $$cod(N)\leq cod(N_1)+cod(N_2)$$ (consider the mapping $\xi\mapsto \langle\bar{\xi_1},\bar{\xi_2}\rangle$ when $\bar{\xi_i}$ is the coset of $N_i$ containing $\xi$)
Question
This is the question I am struggling with:
2.11 In the above exercise, suppose that $cod(N_1)=cod(N_2)$, that is, $d(V/N_1)=d(V/N_2)$.
Prove that $d(N_1/N)=d(N_2/N)$
My attempt
I could only prove it when $V$ is finite-dimensional.
Let $\pi_1\in Hom(N_1, N_1/N)$ and $\pi_2\in Hom(N_2, N_2/N)$ be both the mapping $\xi\mapsto N+\xi$.
$N$ is the null space of $\pi_1$ and $\pi_2$ therefore $\pi_1|_{N_1\setminus N\cup \{0\}}$ and $\pi_2|_{N_2\setminus N\cup \{0\}}$ are isomorphisms.
I can now use a theorem from the book:
Theorem 5.3. Let $T$ be a surjective linear map from the vector space $V$
to the vector space $W$, and let $N$ be its null space. Then a subspace $M$ is a
complement of $N$ if and only if the restriction of $T$ to $M$ is an isomorphism
from $M$ to $W$.
Hence, $N_1=N\oplus (N_1\setminus N \cup \{0\})$ and $N_2=N\oplus (N_2\setminus N \cup \{0\})$.
It is given that $cod(N_1)=cod(N_2)$ therefore I can apply exercise 2.9 and get two subspaces $M_1,M_2$ such that $d(M_1)=d(M_2)$ and $V=N_1\oplus M_1=N_2\oplus M_2$
I can now substiture and get $V=N\oplus (N_1\setminus N \cup \{0\})\oplus M_1$ and $V=N\oplus (N_2\setminus N \cup \{0\})\oplus M_2$
If $V$ is finite dimensional then $$d(N)+d(N_1\setminus N \cup \{0\})+d(M_1)=d(V)=d(N)+d(N_2\setminus N \cup \{0\})+d(M_2)$$
$$\Rightarrow d(N_1\setminus N \cup \{0\})=d(N_2\setminus N \cup \{0\})$$
But $N_1\setminus N \cup \{0\}$ is isomorphic to $N_1/N$ and $N_2\setminus N \cup \{0\}$ is isomorphic to $N_2/N$
Therefore $d(N_1/N)=d(N_2/N)$
(d(V) is the dimension of V)
If we use the concept of quotient-spaces, we can reason like this. Since $V/N_1\cong (V/N)/(N_1/N)$ and $\operatorname{dim}(V/N)<\infty$ it follows that $$ \operatorname{dim}(V/N_1)=\operatorname{dim}(V/N)-\operatorname{dim}(N_1/N). $$ Similarly $$ \operatorname{dim}(V/N_2)=\operatorname{dim}(V/N)-\operatorname{dim}(N_2/N). $$ Since by the assumption $\operatorname{dim}(V/N_1)=\operatorname{dim}(V/N_2)$, then $\operatorname{dim}(N_1/N)=\operatorname{dim}(N_2/N)$.
Note that it does not matter whether $V$ is finite dimensional or not.
At the request of the author of the question @Freud, I give a proof of the formula $$ \dim(V/N_1)=\dim(V/N)-\dim(N_1/N). $$ Let's denote for brevity $\overline{V}=V/N$, $\overline{N}_1=N_1/N$. In view of the formula $$ V/N_1\cong (V/N)/(N_1/N), $$ it is sufficient to check that $$ \dim(\overline{V}/\overline{N}_1)=\dim(\overline{V})-\dim(\overline{N}_1).\hspace{120pt}(1) $$ We know that $\overline{V}$ has finite dimension and $\overline{N}_1<\overline{V}$. Take $\{\bar{e}_1,\ldots,\bar{e}_s\}$ a basis of $\overline{N}_1$ and extend it to $$ \{\bar{e}_1,\ldots,\bar{e}_s,\bar{e}_{s+1},\ldots,\bar{e}_t\} $$ a basis of $\overline{V}$. Then $$ \{\bar{e}_{s+1}+\overline{N}_1,\ldots,\bar{e}_t+\overline{N}_1\} $$ form a basis for $\overline{V}/\overline{N}_1$.
We have $$ \dim(\overline{N}_1)=s,\ \dim(\overline{V})=s+t,\ \dim(\overline{V}/\overline{N}_1)=t. $$ Hence our formula $(1)$ follows.