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?

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