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Suppose $f\colon X \rightarrow Y$ is a continuous map of topological spaces and $s\colon Y \rightarrow X$ is a continuous section to $f$, i.e., $f\circ s = 1$. If $f$ is proper does this mean that $s$ is proper as well? (A continuous map is proper if the preimage of any compact set is compact.)

This is true for schemes and I was wondering whether the same is true for topological spaces.

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The preimage under $s$ of a compact set $K\subseteq X$ is contained in $f(K)$, which is compact. If compact subsets of $X$ are closed, then the answer is then yes. – Mariano Suárez-Alvarez Oct 4 '11 at 2:59
Sure. But I'm interested in the general situation with no assumptions on X and Y. – Unknown Oct 4 '11 at 3:05
Schemes are separated... – Mariano Suárez-Alvarez Oct 4 '11 at 3:06
Dear Mariano, since 1971 schemes are no longer separated. In the original edition of EGA I, published by the IHES in 1960, schemes were indeed defined as separated preschemes. However a new edition of EGA I was published (by Springer) in 1971, in which the old "preschemes" were renamed "schemes" and the old "schemes" were renamed "separated schemes". So that since this date the word "scheme" implies no separation property. To my knowledge all books and articles published after 1971 have adopted the revised terminology, and consequently the word "prescheme" is no longer in use. – Georges Elencwajg Oct 4 '11 at 10:10
Right! ${}{}{}$ – Mariano Suárez-Alvarez Oct 8 '11 at 1:53
up vote 7 down vote accepted

Not in general. As Mariano stated in the comments, it's true if $X$ is $T_2$. Here's a counterexample when $X$ is only $T_1$:

Let $X$ be the space obtained by gluing two closed unit intervals (call them $I_1,I_2$) along some open subinterval. Let $Y$ be a closed unit interval, $f:X\rightarrow Y$ the obvious quotient map, and $s$ one of the two obvious lifts, say to $I_1$. Since $X$ is compact and $Y$ is Hausdorff, $f$ is proper. However, the preimage of $I_2$ in $Y$ will be an open interval, so $s$ is not proper.

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