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$X$ is compact and $f$ :$X$ $ \to $ $Y$ is continuous and onto.

Then show that $f$ is an open map. i.e. Every open set of $X$ is mapped to an open set of $Y$.

Shreedhar Bhat
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1 Answers1

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As the question was first stated (without "onto"): Not even close to the truth. Consider $Y$ a non-discrete space, $c\in Y$ a point such that $\{c\}$ is not open, $X$ any topological space and $f:X\to Y$ the constant map $f\equiv c$. Any non-empty open subset of $X$ is mapped to a non-open set.

What if it is onto: Still no. Consider $X=[0,10^4]$, $f(x)=x\sin x$ and $Y=[a,b]=f([0,10^4])$ the image interval. $f(x)$ has a sequence $x_1\le y_1\le x_2\le y_2\le x_3\le y_3\le\ldots$ of local maxima $x_n$ and local minima $y_n$ such that $f(x_n)\nearrow +\infty$ and $f(y_n)\searrow-\infty$. For a fixed $n$ such that - say - $y_{2n}< 10^4$, and a sufficiently small $\delta$, $f((x_n-\delta,y_n+\delta))=[f(y_n),f(x_n)]\subseteq (a,b)$. Therefore, $[f(y_n),f(x_n)]$ is not open.

What if $f$ is onto and closed, and $Y$ Hausdorff? Consider the map $\operatorname{cis}:[-\pi,\pi]\to S^1$, where $\operatorname{cis}(x)=e^{ix}$. Since it is a continuous map from a compact space to a $T_2$ topological space, it is closed. However, $$\operatorname{cis}((0,\pi])=\{z\in S^1\,:\, \Im z\ge0\}\setminus\{1\}$$ which is not open.

What if it is bijective but $Y$ is not Hausdorff? (because yeah): The first example can be adapted. Consider a finite topological space $(X,\tau)$ such that, for at least one $c\in X$, $\{c\}$ is not open (for instance, $\#X\ge 2$ and $\tau=\{\emptyset,X\}$). If $\mathcal P(X)$ is the discrete topology on $X$, the map $id:(X,\mathcal P(X))\to (X,\tau)$ is always continuous. Since $X$ is finite, $(X,\mathcal P(X))$ is compact and all subsets are open. However, by hypothesis $id(\{c\})=\{c\}\notin\tau$.

  • What if the mapping was onto, is it true then ? – Shreedhar Bhat May 10 '17 at 18:46
  • Let $C$ be closed in $X$. We have $Y=f(X)$ is compact. We therefore have $f(C)$ is compact. If we assume $Y$ is Hausdorff then $f(C)$ is closed and hence f is an open mapping. –  May 10 '17 at 18:54
  • @SoumikGhosh By "hence $f$ is an open mapping" I think you mean "hence $f$ is a closed mapping"? Of course, if $f$ is a bijection, then it is also an open mapping. – bof May 10 '17 at 19:19
  • yes that's what i meant for onto maps closed mappings are open mappings –  May 10 '17 at 19:27
  • @SoumikGhosh That's where you are wrong: onto closed maps may not be open. –  May 10 '17 at 19:29
  • suppose f is a closed onto map. Let U be open in X then f($U^c$) is closed and f(U)=Y\f($U^c$)which is open –  May 10 '17 at 19:35
  • @SoumikGhosh The map $[-\pi,\pi]\ni x\mapsto \exp(ix)\in S^1$ is continuous, closed and onto (it's even a quotient map), but it is not open. –  May 10 '17 at 19:45
  • oh i assumed injectivity my mistake –  May 10 '17 at 19:47