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Dual image map restricts to open sets?

Mathematics Asked by blargoner on September 18, 2020

A book I’m reading on category theory says that if $A$ and $B$ are topological spaces and $f:Ato B$ is continuous, then the "dual image" map
$$f_*(U)={,bin Bmid f^{-1}(b)subseteq U,}$$
restricts to open sets; that is, $f_*:mathcal{O}(A)tomathcal{O}(B)$. (So then it’s right adjoint to $f^{-1}:mathcal{O}(B)tomathcal{O}(A)$.)

This seems wrong, since it would imply for example (taking $U=varnothing$) that the image of a continuous function is always closed.

Are there natural conditions under which it does make sense to restrict to open sets?

One Answer

You’re right that there’s a problem here.

Let $A=(0,1)times(0,1)$ and $B=(0,1)$, each with the usual topology, and let $$f:Ato B:langle x,yranglemapsto x$$ be the projection to the $x$-axis; this is a continuous, open map. Let $$U={langle x,yranglein A:y>2x-1};.$$ $U$ is open in $A$, but

$$f_*(U)=left(0,frac12right];,$$

which is not open in $B$.

You do get the result if $f$ is closed and continuous. In that case let $U$ be open in $A$, and let $F=Asetminus U$. Suppose that $bin B$; then $f^{-1}[{b}]subseteq U$ iff $bnotin f[F]$, i.e., iff $bin Bsetminus f[F]$, so $f_*(U)=Bsetminus f[F]$, which is open in $B$.

Correct answer by Brian M. Scott on September 18, 2020

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