# Example of a compact space in $\mathbb R^3$ which is contractible but doesn't deformation retract to any point

Update:

I think of a way that $$Y$$ can defromation retract to $$D^2$$ in the weak sence, so from Exercise $$0.4$$, $$Y$$ is contractible. See tha answer below.

This is Exercise $$7$$ of chapter $$0$$ in page $$18$$ of Hatcher's Algebraic Topology.

The following explanation is from Hatcher himself:

$$X$$ is union of an infinite sequence of cones on the Cantor set arranged end-to-end, and $$[0,1)$$ is the "baseline". Y is one-point compactification of $$X \times \mathbb R$$, which is obtained by adding the endpoint $$1$$ of the interval $$[0,1)$$.

My question:

Why is $$Y$$ contractible?

This question was asked before: Hatcher Chapter $$0$$ Exercise $$7$$ , but the answer might contain some mistake. It says "The space $$Y$$ is contractible, since we can retract each of the fins and obtain the closed disc $$D^2$$, which is contractible."

Some clarification: The picture in Hatcher's book is equivalent to the original construction from Edwards, which is stated in the answer.

But only retraction doesn't preserve the homotopy type, retracting each of the fins is not deformation retraction since it's not continuous, and $$Y$$ doesn't deformation retract to $$D^2$$.

Thanks for your time and effort!

• You can stick the word "deformation" in front of "retract", to make that sentence more clear: The space $Y$ is contractible, since we can deformation retract each of the fins... Jan 15, 2020 at 23:00
• @LeeMosher That's not a true deformation retraction actually, just a deformation retraction in the weak sence. Jan 16, 2020 at 15:08

$$X$$ is union of infinite sequence of cones on the Cantor set arranged end-to-end and getting smaller and smaller. The ''baseline'' of $$X$$ is $$[0,1)$$. One-point compactification of $$X \times \mathbb R$$ is obtained by adding the endpoint $$1$$ of $$[0,1)$$.

After one-point compactification, $$\{0\}\times \mathbb R$$ together with the additional point $$\{1\}\times \{0\}$$ becomes the boundary of $$D^2$$.

$$Y$$ is obtained from one-point compactification of $$X \times \mathbb R$$ by wrapping one more cone on the Cantor set around the boundary of $$D^2$$.

$$Y$$ doesn't deformation retract to a point, since any neighborhood of a point in $$Y$$ is disconnected.

$$X$$ can deformation retract to baseline $$[0,1)$$ in the weak sense in the following way:

For $$n \in \mathbb N$$, the point on baseline $$[1-\frac{1}{2^{n+1}}, 1-\frac{1}{2^{n+2}}]$$ moves to $$[1-\frac{1}{2^{n}}, 1-\frac{1}{2^{n+1}}]$$ alone $$[0,1)$$, and the point on $$[0, \frac{1}{2}]$$ moves to $$\{0\}$$.

The point on cones moves to $$[0,1)$$ in the similar way, so $$X$$ deformation retract to baseline $$[0,1)$$ in the weak sense, and one-point compactification of $$X \times \mathbb R$$ deformation retract to $$D^2$$ in the weak sense.

$$Y$$ deformation retract to $$D^2$$ with a cone on the Cantor set around the boundary of $$D^2$$ in the weak sense.

This space can deformation retract to $$D^2$$ in the weak sense by moving points on cone and rotating $$D^2$$ clockwise.

Thus $$D^2 \hookrightarrow Y$$ is homotopy equivalence, $$D^2$$ is contractible, so $$Y$$ is contractible.