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A topological space $X$ is called compactly generated if following condition holds:

$A\subseteq X$ is open in $X$ iff for every compact $K\subseteq X$, $A\cap K$ is open in $K$.

My lecturer said that a topological space $X$ is called compactly generated if following condition holds:

$A\subseteq X$ is open in $X$ iff for every compact $K$ and continuous map $f: K\rightarrow X, f^{-1}(A)$ is open in $K$.

the necessary condition is clear. To prove the sufficient, I bit confuse. I think if $f: K\rightarrow X$ is continuous and $f^{-1}(A)$ is open in $K$, obviously that $A$ is open in $X$. Hence we no need condition $X$ to be a compactly generated.

Please tell me if it is not true.

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  • $\begingroup$ Note that , the formal statements stats that if $f$ is continuous and A is open , then $f^{-1}(A)$ is open,thus if $f^{-1}(A)$ is open and $f$ is continuous this does not imply that A is open. $\endgroup$ – Nizar May 14 '15 at 8:00
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The definitions are equivalent because, for a set $A\subseteq X$, the intersection $A\cap K$ is open in $K$ for every compact set $K\subseteq X$ if and only if $f^{-1}(A)$ is open in $L$ for every map $f$ from a compact space $L$ to $X$.

For one direction you need the fact that the image of a compact space under a continuous map is compact. For the other direction try to express the intersection $A\cap K$ as the preimage under some map $i:K\to X$.

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  • $\begingroup$ I did not understand your answer. For one direction is obvious. But for another direction is not clear. $\endgroup$ – lili May 14 '15 at 14:36
  • $\begingroup$ @lili: Could you say which direction is unclear? $\endgroup$ – Stefan Hamcke May 14 '15 at 15:24
  • $\begingroup$ if $f^{-1}$ is open in $K$ ...... then $A$ is open in $X$. Should $K$ compact subset of $X$? $\endgroup$ – lili May 14 '15 at 15:30
  • $\begingroup$ @lili: So assume $f^{-1}(A)$ is open in $K$ for every map $f:K\to X$ from a compact space $K$. Now if $K\subseteq X$ is compact, the inclusion $i:K\to X, x\mapsto x$ is such a map, so by assumption, $i^{-1}(A)=A\cap K$ is open in $K$. $\endgroup$ – Stefan Hamcke May 14 '15 at 15:40
  • $\begingroup$ I know it. But I am thinking whether $K$ is any compact space (it no need to be subset of $X$), does $A$ still open in $X$? $\endgroup$ – lili May 14 '15 at 15:45

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