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Let $X$ be a topological space containing a point whose closure is the whole space $X$. Then is $X$ contractible ?

I feel it is, but I am unable to come up with a proof. Please help.

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If $\eta$ is the dense point and $I=[0,1]$ the map $f:X\times I\to X$ defined by $$f(x,0)=x \quad \operatorname {and} \quad f(x,t)=\eta \quad \operatorname {for} t\gt0$$ is the required contraction of $X$ to $\eta$ .

Georges Elencwajg
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    Just out of curiosity: this implies that every integral scheme is contractible, right? Because the generic point is dense. This seems very strange indeed: because then homology and cohomology in positive degree should be trivial. Or do we only define homology and cohomology on, say, complex points? – 57Jimmy Mar 21 '18 at 15:01
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    @57Jimmy 3: Yes, every integral, or even irreducible, scheme is contractible. However the Zariski topology is never used for calculating homology, cohomology or fundamental group of a scheme. As you correctly suggest, for a scheme $X$ over $Spec (\mathbb C)$ one only studies the complex points $X(\mathbb C)$ and moreover one endows those with the classical topology deduced from the metric topology of $\mathbb C$. Key words: GAGA, Hodge. Or, in the arithmetic case, one endows the scheme with its powerful étale topology, which is alas much more complicated than a topology in the usual sense. – Georges Elencwajg Mar 21 '18 at 18:24
  • Great, thanks for your answer! – 57Jimmy Mar 22 '18 at 11:50
  • You are welcome, @57Jimmy. – Georges Elencwajg Mar 22 '18 at 12:00