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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.

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  • $\begingroup$ I asked this very question here. $\endgroup$ – Tyrone Apr 29 '20 at 10:04
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    $\begingroup$ @Tyrone Your question asks for a space in which no point is a strong deformation retract. The OP considers deformation retracts. $\endgroup$ – Paul Frost Apr 30 '20 at 11:17
  • $\begingroup$ @PaulFrost thanks for paying attention. Please feel free to replace ' this very' with 'a similar' in the above. ;) $\endgroup$ – Tyrone Apr 30 '20 at 13:10
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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 $F(a,1) = a$.

An example is the comb space (see the above link). A more sophisticated example is given in the answer to Tyrone's question Contractible space which is not pointed-contractible. At all. Ever. Under any circumstances. Here no point is a strong deformation retract of the space.

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.

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