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Let $X$ and $Y$ be metric spaces. Let us say that $f\colon X\to Y$ is uniformly open if for every $\varepsilon >0$ there is $\delta > 0$ such that for any $x\in X$ $$ B(f(x),\delta) \subseteq f[B(x,\varepsilon)].$$

(See p. 202 of Kelley's General topology.)

It is not too hard to show that if $X,Y,Z$ are metric spaces and $f\colon X\to Y$ is a uniformly continuous surjection, $g\colon Y\to Z$ is a function such that $g\circ f$ is uniformly open, then $g$ is uniformly open. (A more abstract version of this statement can be found here - Proposition 1.13.)

Consider $X=\mathbb C \times \mathbb C$, $Y=Z=\mathbb C$ and $f(x,y)=x+y$ and $g(z)=e^z$. Then $f$ is a uniformly continuous surjection. Let us look at the composition $g\circ f$. This is nothing but multiplication in $\mathbb C$ restricted to $\mathbb{C}\setminus \{0\}$. However, multipliction in $\mathbb C$ is uniformly open with $\delta(\varepsilon)=\varepsilon^2/4$. Thus $g$ should be uniformly open but it is not.

What is wrong here?

user512365
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1 Answers1

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When one considers the map $g\circ f:(x,y)\mapsto e^{x+y}$ as multiplication in $\mathbb{C}$ restricted to $\mathbb{C}^*:=\mathbb{C}\backslash \{0\}$, one is precomposing it with the map $(x',y')\mapsto (\log\,x',\log\,y')$ which really alters any property that $g\circ f$ could have before precomposing it with that map.

In other words, $(x,y)\mapsto e^{x+y}$ and $(u,v)\mapsto uv$ are radically different maps even though we can see them as the same in some special contexts, but that requires a justification and it will fail when dealing with metrical/uniform properties of a map.

Josué Tonelli-Cueto
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