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The setting:

  1. $X$ and $Y$ are topological vector spaces.
  2. $N \subset X$ is a closed subspace.
  3. $T(N)=\{0\}$
  4. $\pi : X \rightarrow X/N$ the quotient map.
  5. $S : X/N \rightarrow Y$ uniquely determined by $T=S \circ \pi$ ($S$ is well-defined by $S(x+N)=T(x)$ and is linear).

The homework:

  1. $T$ is continuous iff $S$ is continuous.
  2. $T$ is open iff $S$ is open.

I "proved" both statements, but I guess I have a mistake because I never used the fact that $N$ is closed.

My attempt:

  1. If $T$ is continuous and $U \subset Y$ is open than $\pi^{-1}(S^{-1}(U))=T^{-1}(U)$ is open because $T$ is continous.
  2. I showed that $\pi$ is open and thus if $S$ is open, so is $T=S \circ \pi$. As for the other direction, I showed that if $B\subset X/N$ is open then $S(B)=T(A)$ for some $A \subset X$ such that $A+N$ is open. So $S(B)=T(A)=T(A+N)$ is open if $T$ is open.

Is it really necessary that $N$ is closed?

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1 Answer 1

up vote 0 down vote accepted

For (1), I would just add that, because $\pi^{-1}(S^{-1}(U))=T^{-1}(U)$ is open, $S^{-1}(U)$ is open in $X/N$ by the definition of the quotient topology.

For (2), if $A=\pi^{-1}(B)$, then $A$ is open and $\pi(A)=B$, so $S(B)=S(\pi(A))=T(A)$ is open if $T$ is open. Other than that, you're argument for (2) is also correct. My guess is that the hypothesis that $N$ is closed is imposed because topological vector spaces are usually required to be Hausdorff (at least over $\mathbf{R}$ or $\mathbf{C}$), and if $N$ is not closed, then $X/N$ won't be Hausdorff.

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