# Give an example of a contractible space X which has no deformation retraction to any point of X

I am new in topological algebra however, I know an example of a space X which has just one point $$x_{0}\ \in\ X$$ that $$x_{0}$$ is a deformation retraction of X:

A cone CQ wich Q is rational number set, is just have one deformation retract point and its the vertex point $$x_{0}$$.

deformation retraction:

A continuous map

$$F:X \times [0,1] \rightarrow X$$

is a deformation retraction of a space X onto a subspace A if, for every x in X and a in A,

$$F(x,0)=x,\ F(x,1)\ \in \ A\ and\ F(a,1)=a$$

The subspace A is called a deformation retract of X.

Contractible space

a topological space X is contractible if the identity map on X is null-homotopic, i.e. if it is homotopic to some constant map. Intuitively, a contractible space is one that can be continuously shrunk to a point within that space.

my question is: I need an example of a topological space which is contractible but has no point to be its deformation retraction.

• I asked this very question here. – Tyrone Apr 29 '20 at 10:04
• @Tyrone Your question asks for a space in which no point is a strong deformation retract. The OP considers deformation retracts. – Paul Frost Apr 30 '20 at 11:17
• @PaulFrost thanks for paying attention. Please feel free to replace ' this very' with 'a similar' in the above. ;) – Tyrone Apr 30 '20 at 13:10

A contractible space $$X$$ has each point as a deformation retract. In my answer to Is Armstrong saying that the comb space is not contractible? you can find a proof that $$X$$ is contractible to any $$x_0 \in X$$. This means that there exists a contraction of $$X$$ to $$x_0$$, i.e. a homotopy $$F :X \times I \to X$$ such that $$F(x,0) = x$$ and $$F(x,1) = x_0$$ for all $$x \in X$$. This $$F$$ is a deformation retraction of $$X$$ onto $$\{x_0\}$$ as defined in your question.
However, in general $$X$$ does not have each point $$x_0$$ as a strong deformation retract. Recall that $$A \subset X$$ is a strong deformation retract of $$X$$ if there exists a strong deformation retraction $$F : X \times I \to X$$ such that $$F(x,0) = x, F(x,1) \in A$$ for all $$x \in X$$ and $$F(a,t) = a$$ for all $$a \in A$$ and $$t \in I$$. This strengthens the definition of a deformation retraction by requiring that $$F$$ keeps all points of $$A$$ fixed. This is not required for a deformation retraction - it only requires $$F(a,1) = a$$.
Also note that $$x_0$$ is a strong deformation retract of $$X$$ if and only if $$X$$ is pointed contractible to $$x_0$$ which means that there is a contraction $$F : X \times I \to X$$ which keeps $$x_0$$ fixed.