If $X$ is the retract of its cone $CX$ it is contractible

Question

I'm trying to prove this statement, but find that "retract" only implies a continuous map from the cone of the underlying space to itself but doesn't offer information about contractibility of the underlying space. Hence how to construct the homotopy from the identical map to the constant one ?

Answer 1

If you write out all the "data" that you have, the contracting homotopy almost jumps out at you. To be definite, we will define the cone $CX$ of $X$ as the quotient of $X \times [0,1]$ by gluing the top of the cylinder together: $(x,1) \sim (x',1)$ for all $x, x' \in X$ will be our relation. Note that this means we have a canonical quotient map $q: X \times [0,1] \to CX$ which necessarily satisfies $q(x,1) = q(x',1)$ for all $x, x' \in X$. In other words, $q$ is a constant on $X \times \{1\}$. Hmmmm... that sounds like it may be contracting something.

Write out what you have: $X \times I \stackrel{q}{\rightarrow} CX \stackrel{r}{\rightarrow} X$, where $r: CX \to X$ is your retraction. (Here, $X$ is embedded as the bottom of the cylinder $X \times \{0\}$ pre-identifications.) This map is your homotopy. So, let $H=r \circ q$. Note that $$ H(x,0) = rq(x,0) = r(x,0)= (x,0) $$ because $r$ is a retraction. On the other hand, by the first paragraph, $$ H(x,1) = rq(x,1)=r(\text{constant}) = \text{constant}. $$ Hence $H$ is a homotopy between the identity and a constant, so $X \cong X \times \{0\}$ is contractible.

Answer 2

The cone $CX$ is the quotient of $X \times [0, 1]$ under the relation that contracts the subspace $X \times \{1\}$ to a point. $X$ is embedded into $CX$ as the subspace $X \times \{0\}$. If $X$ viewed as a subspace of $CX$ is a retract of $CX$, then the projection of $X \times [0, 1]$ onto $CX$ composed with the retraction mapping $CX \to X$ is a homotopy from the identity on $X$ to a constant function.

Answer 3

Here is an approach using abstract nonsense. If you know that cones are contractible, you can do this by proving that a retract of contractible space is contractible:

Consider $hTop$ the category of topological spaces whose morphisms are the homotopy classes of continuous maps. If a space $Z$ is terminal in $hTop$ then the only arrow in the homset $[Z,Z]$ is the homotopy class of the identity But, we $always$ have the constant maps from $Z$ to $Z$. Thus, $1_Z\sim c$ where $c$ is any constant map. i.e. $Z$ is contractible.

On the other hand, if a space $Z$ is contractible, then there is a homotopy equivalence from $Z$ to a singleton, $\left \{ * \right \}.$ This means that $Z\cong \left \{ * \right \}$ and since $\left \{ * \right \}$ is terminal in $hTop$, so is $Z$. i.e. for any space $Y$, the homset $[Y,Z]$ consists of a single arrow. In our case, $[Y,CX]$ consists of a single arrow.

Let $\phi:Y\to X$. Then, if $i$ is the inclusion: $X\to CX$, we have $i\circ \phi\in [Y,CX]$ and as $CX$ is terminal, this is the only arrow in $[Y,CX]$, which we may call $\psi$. Then, $\phi=r\circ i\circ\phi =r\circ \psi,\ $ so $\phi$ is uniquely determined; that is to say, the homset $[Y,X]$ consists of a single arrow, which means that $X$ is terminal in $hTop$ and therefore contractible.