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Suppose $f:X\longrightarrow Y$ is a continuous map between topological spaces. By definition, this means that if $V$ is open (in $Y$), then $f^*(V)$ is open (in $X$), where $f^*$ denotes preimage.

I wonder if the following condition is equivalent to continuity.

  • Property 1. If $f(U)$ is open (in $Y$), then $U$ is open (in $X$).

Now suppose $f$ satisfies property 1. I think the preimage of $f(U)$ is $U$, because $f(x)$ is in $f(U)$ iff $x$ is in $U$. So the preimage of the open set $f(U)$, being $U$, is indeed open, because $U$ is open by property 1; this is what you'd expect if $f$ is continuous. So continuity should at least imply property 1.

Martin Sleziak
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étale-cohomology
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1 Answers1

5

Let $f\colon \mathbb{R} \to \mathbb{R}$ send negative numbers to 0 and nonnegative reals to 1. The image of $f$ contains no open sets, so condition 1 is trivially true, but $f$ is not continuous, so the property does not imply continuity.

For a counterexample the other way, take for example the continuous map $x \to |x|$ on the real numbers. Let $U$ be $(0,1/2) \cup (1/2,1) \cup \{-1/2\}$ So we have taken an open interval, and replaced a single point with another far away point with the same image. Then $f(U)$ is the interval $(0,1)$ which is open, but $U$ is not open.

Carl
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    Before reading your A, I was going to give $f:\Bbb R\to \Bbb R$ where $f(x)=x-1$ for $x<0$ and $f(0)=0$ and$ f(x)=x+1$ for $x>0$ but your example is simpler, – DanielWainfleet Dec 12 '17 at 08:34