# Preimage of a closed set under surjection

Let $X,Y$ be topological vector spaces, $\overline{B(0,1)}\subset Y$ is compact, and $T:X\to Y$ is a surjection. Would like to show that $W\subset Y$ is closed then $T^{-1}(W)\subset X$ is closed.

My attemp is, since topological vector spaces are Hausdorff, then $X,Y$ are Hausdorff. And since compact subspace of hausdorff space is closed, hence $\overline{B(0,1)}\subset Y$ is closed. Since $T$ is a surjection then $T^{-1}(\overline{B(0,1)})\subset X$. Now it remains to show that $T^{-1}(\overline{B(0,1)})\subset X$ is closed. But I do not know how to do that. I know that $T^{-1}(\overline{B(0,1)})$ is convex and balanced, but I am not sure whether this information useful. Note that I do not say that $T$ is continuous. Any help is appreciated! :)

• Is $T$ supposed to be linear? And what is $\overline{B(0,1)}$? – Eric Wofsey Aug 14 '16 at 5:03
• @EricWofsey yes, $T$ is linear transformation. $\overline{B(0,1)}$ is a closure of open ball of radius $1$ centered at $0$. – Chen M Ling Aug 14 '16 at 5:05
• What is an "open ball of radius $1$"? Are you assuming $Y$ is a normed space? – Eric Wofsey Aug 14 '16 at 5:05
• Basically it is a neighborhood of $0$ which is compact – Chen M Ling Aug 14 '16 at 5:12

This isn't true. For instance, take $Y=\mathbb{R}$ and $X=\ell^1$ (in fact, this can be modified to work with $X$ as any infinite-dimensional Banach space). Let $Z\subset X$ be the subspace consisting of sequences that are eventually $0$; note that $Z$ is dense in $X$. Let $f:X/Z\to\mathbb{R}$ be a linear surjection and compose it with the quotient map $X\to X/Z$ to get a linear surjection $T:X\to\mathbb{R}$. Then $T^{-1}(\{0\})$ contains $Z$ and so cannot be closed.
• What is $\ell^1$ space? – Chen M Ling Aug 14 '16 at 5:15
• The space of sequences $(x_n)$ such that $\sum |x_n|<\infty$, with $\|(x_n)\|=\sum|x_n|$ as the norm. That is, it is the $L^1$ space for $\mathbb{N}$ with counting measure. – Eric Wofsey Aug 14 '16 at 5:17
• The $L^p$ spaces are infinite dimensional, so it contradicts with the fact that $Y$ has compact neighborhood of $0$. – Chen M Ling Aug 14 '16 at 5:22
• $Y$ is just $\mathbb{R}$ here; it's $X$ that's infinite-dimensional. – Eric Wofsey Aug 14 '16 at 5:34