# Quotient maps: If $X$ is locally convex space, then so is $X/N$.

So I am having trouble understanding quotient map in my functional analysis class. We are given the following exercise for practice. let $$X$$ be a topological vector space and let $$N\subseteq X$$ be a closed subspace. Take $$Q:X\rightarrow X/N$$ to be the quotient map. I've managed to show that if $$p$$ is a semi-norm on $$X$$, then $$p_N([x])=inf\{p(x+y):y\in N\}$$ is a semi-norm on $$X/N$$. The next step is to show that the $$Q(\{y\in X:p(x-y), which I am having issues with. I just can't get my head around what the LHS looks like. The next part of the exercise is to show that the space $$X$$ being a locally convex space implies that $$X/N$$ is as well. Any help with that part would be appreciated as well. Thanks in advance.

• What do you know about the properties of $Q$? Do you know that $Q$ is open? In fact, $Q$ maps the open unit ball in $X$ to the open unit ball in $X/N.$ – Matematleta Jan 25 '19 at 1:22
• I've proved that $Q$ is open, but I was unaware of the other fact. Additionally, I know that $Q$ is continuous and linear. – Scott Jan 25 '19 at 1:25
• OK, if you know that $Q$ is open, the proof is not too bad. See my answer. – Matematleta Jan 25 '19 at 1:57

For the first part, wlog $$x=0$$, and let $$B_r=\{y\in X:p(y) and $$B_r^N=\{[y]\in X/N:p_N([y]).
If $$y\in B_r$$ then by definition of $$p_N$$, we have $$p_N([y])\le p(y) so $$Q(B_r)\subseteq B^N_r.$$
On the other hand, if $$[y]\in B_r^N$$, then choose $$n\in N$$ such that $$p(y+n). Then, $$y+N=y+n+N=Q(y+n)\Rightarrow B_r^N\subseteq Q(B_r).$$ The result follows.
For the second part, let $$Q : X → X /N$$ be the quotient map. Choose a neighborhood $$V$$ of $$0$$ in $$X /N$$. Since $$V$$ is a neighborhood of $$0$$ and $$Q$$ is continuous, the pre-image $$Q^{-1}(V)$$ is a neighborhood of $$0$$ in $$X$$. Now, since $$X$$ is locally convex, there exists an open convex set $$U \subseteq X$$ , with $$0 \in U \subseteq Q^{-1}(V)$$. Then, $$Q(U)\subseteq V$$ is open (since $$Q$$ is open). The fact that $$Q(U)$$ is also convex is almost immediate:
suppose $$[x], [y]\in Q(U)$$ and consider $$t[x]+(1-t)[y];\ 0\le t\le 1$$. Expanding, we get $$t(x+N)+(1-t)y+N=tx+(1-t)y+N$$ and now the result follows since $$tx+(1-t)y\in U$$ (since $$U$$ is convex), which means that $$t[x] +(1-t)[y]=\in Q(U).$$
• How does this show that $Q(\{y\in X:p(x−y)<r\})=\{[y]\in X/N:p_N([x]−[y])<r\}$? – Scott Jan 25 '19 at 3:10