Let $X$ be a topological vector space (not necessarily Hausdorff), with topology $\tau$, and $M, N$ linear subspaces of $X$. Let $\pi:X \rightarrow X/N$ be the quotient map, which associates to each $x \in X$ the coset $x + N \in X/N$, and consider the linear subspace $\pi(M)$ of $X/N$.


Define the map $\Phi: M /(M \cap N) \rightarrow \pi(M)$ by associating to each coset $x+ (M \cap N)$, with $x \in M$, the coset $x+N$. It is immediate to see that $\Phi$ is well defined (that is $\phi(x+(M \cap N))$ does not depend on the particular $x$ chosen in the coset $x+ (M \cap N)$) and that it is an isomorphism of vector spaces.


Moreover, in Note (1) below, it is shown that $\Phi$ is continuous. Anyway, in general $\Phi$ is not a homeomorphism: in Note (3) below a counterexample is given in which $X$ is Hausdorff and $M$ is closed.
Can you find a counterexample in which $X$ is Hausdorff and both $M$ and $N$ are closed?


Thank you very much in advance for your attention.


NOTE (1) Suppose that $V$ is open in $\pi(M)$, that is $V=E \cap \pi(M)$, with $\pi^{-1}(E)=W \in \tau$. Then $\Phi^{-1}(V)= \{ x + (M \cap N) \in M/(M \cap N): x + N \in V \}$. So, if $\rho : M \rightarrow M/(M \cap N)$ is the quotient map (which associates to $x \in M$ the coset $x + (M \cap N)$), then we have 
\begin{equation}
\rho^{-1}(\Phi^{-1}(V))= \{ x \in M : x+N \in V \}= M \cap W.
\end{equation}
 So $\rho^{-1}(\Phi^{-1}(V))$ is an open set of $M$, and we conclude by definition of quotient topology that $\Phi^{-1}(V)$ is an open set of $M/(M\cap N)$.


NOTE (2) If $N \subset M$, then $\Phi$ is open, so it is a homeomorphism. Indeed in this case $M \cap N = N$. Let $F \subset M/N$, with $\rho^{-1}(F)=M \cap W$ and $W \in \tau$. Put $E= \pi(W)=\pi(W+N)$. We have $\pi^{-1}(E)= W + N \in \tau$, so $E$ is an open set of $X/N$. 


Note that we have $M \cap W =M \cap (W + N)$. Indeed if $x \in M \cap W$, then $x+N \in F$, so that $x + n \in \rho^{-1}(F)=M \cap W$ for all $n \in N$. So 
\begin{equation}
\Phi(F)= F = \{x + N \in M/N: x \in M \cap (W+N) \}.
\end{equation}
If $x+N \in \Phi(F)$, then $x \in M \cap (W + N)$, so that $x+N \in E \cap \pi(M)$. Conversely, if $x+ N \in E \cap \pi(M)$, then for some $m \in M$ and $w \in W$, we have $x+N=m+N=w+N$, so that $m \in M \cap (W+N)$, and $x+N=m+N \in \Phi(F)$. We conclude that $\Phi(F)=E \cap \pi(M)$, which is an open set of $\pi(M)$.


NOTE (3) Let $X$ be a Hausdorff TVS, and $M$ a closed linear subspace which is not complemented (for the definition of "complemented subspace" see Rudin, Functional Analysis, Second Edition, Section 4.20, while for example of uncomplemented subspaces see Rudin, cit., pp. 132-138). Let $\{ v\_{\alpha} \}$ be a Hamel basis of $M$, and let $\{ w\_{\beta} \}$ a subset of $X$ such that $\{ v\_{\alpha} \} \cup \{ w\_{\beta} \}$ is a Hamel basis of $X$. Let $N$ be the linear subspace spanned by the $\{ w\_{\beta} \}$. Then we have $M + N =X$ and $M \cap N = \{0 \}$. Since $M$ is uncomplemented, $N$ cannot be closed. We have $\pi(M)=X/N$ and $M/(M \cap N)= M/\{0 \}$. 


Now, let us remember that if $T$ is a TVS and $S$ a linear subspace of $T$, then the quotient space $T/S$ is Hausdorff if and only if $S$ is closed (see e.g. Horvath, Topological Vector Spaces and Distributions, Proposition (5) at p. 105 or Treves, Topological Vector Spaces, Distributions and Kernels, Proposition (4.5) at p. 34). So $X/N$ is not Hausdorff, while $M/ \{0 \}$ is Hausdorff (since $X$ is Hausdorff, also $M$ is Hausdorff, so $\{0 \}$ is a closed subset of $M$). We conclude that $X/N$ and $M/ \{0 \}$ cannot be homeomorphic. So $\Phi$ cannot be a homeomorphism.



Subspaces of Quotient Spaces
I develop here the suggestion given by Bill Johnson in the comment above. 


Take $X=\ell^2$, and let $M$ be the subspace all the complex sequences $x=(x\_0, x\_1,x\_2, \dots) \in \ell^2$ such that $x\_{2n}=0$ for all non-negative integers $n$, and let $N$ be the subspace of all complex sequences $x=(x\_0, x\_1,x\_2, \dots) \in \ell^2$ such that $x\_{n+1}=nx\_{2n}$ for all non-negative integers $n$ (this example is taken from Robert Israel's answer to the post [The Direct Sum](https://math.stackexchange.com/questions/135471/the-direct-sum-of-two-closed-subspace-is-closed-hilbert-space)).


We prove that $M$ and $N$ are quasi-complements, but not complements. Indeed, clearly $M$ and $N$ are closed linear subspaces, $M \cap N= \{ 0 \}$ and $M+N$ is dense (since $M+N$ contains all sequences which are definitively zero). But $M + N \neq \ell^2$. Indeed $M+N$ does not contain the sequence $x\_n=1/(n+1)$: if it were so, then $x=u+v$, with $u \in M$ and $v \in N$, then $v\_{2n}=1/(2n+1)$ and $v\_{2n+1}=n/(2n+1)$. But then $v \notin \ell^2$. 


Now, let us consider our original question. We shall prove that, with this choice of $X, M$ and $N$, the map $\Phi$ is not a homeomorphism.


Indeed, since $M$ is a closed subspace of $X$, then $M/(M \cap N)= M/\{0 \}=M$ is a sequentially complete TVS. On the other hand, $X/N$ is a Banach space (see e.g. Rudin, Functional Analysis, Second Edition, Theorem (1.41)). Now, if $\Phi$ were a homeomorphism, then $\pi(M)$ and $M$ would isomorphic as topological vector spaces. So $\pi(M)$ would be sequentially complete. But then $\pi(M)$ should be a closed subspace of the Banach space $X/N$. Since $\Phi$ is continuous, we would get that $\pi^{-1}(\pi(M))=M+N$ is closed, which is not the case.


QED


