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Can somebody come up with an example where $X \subset Y$, the inclusion gives a homotopy equivalence between $X$ and $Y$, but there is no deformation retraction from $Y$ onto $X$?

user136592
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    Sure. Take the punctured plane and draw a circle to the right of the puncture. Then the inclusion map is not a homotopy equivalence. –  Jan 24 '16 at 21:36
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    Right, so to make the question interesting you should ask for the inclusion to be a homotopy equivalence. – Qiaochu Yuan Jan 24 '16 at 21:37

2 Answers2

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There are known examples of spaces which are contractible but do not deformation retract onto any point in them. So you can take $Y$ to be such a space and $X$ to be any point in it.

Community
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Qiaochu Yuan
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Consider the space $$Y=\{(x,y) \in \mathbb R^2 \mid \exists m \in \mathbb Q\ y=mx \}$$ this space is contractible, so it is homotopy equivalent to the point $\{(1,0)\}=X$ for instance, nevertheless $\{(1,0)\}$ is not a deformation retract of $Y$.

Giorgio Mossa
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