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The first exercise given in Spanier's, Algebraic Topology is:

$X$ is contractible if and only if it is a retract of any cone over $X$.

I have proven the first implication, however I am stuck on the second implication.

In Spanier he defines a cone over a topological space $X$ with vertex $v$ as the mapping cylinder of the constant map $X\rightarrow v$. In addition he denotes the mapping cylinder as $Z_{f}$.

Starting with the second statement that $X$ is a retract of any cone over $X$, I have shown it is certainly a retract of $Z_{c}$ where c is the constant map from $X$ to $x_{0}$ for some $x_{0}\in X$ and I've also shown that $P$ can be imbedded as a weak deformation retract of $Z_{c}$, where $P$ is the one point space containing $x_{0}$.

Now in Spanier there is a theorem stating:

Two spaces $X$ and $Y$ have the same homotopy type if and only if they can be imbedded as weak deformation retracts of the same space $Z$.

Since I have shown $P$ can be imbedded as weak deformation retract of $Z_{c}$ and since $X$ is already a retract of $Z_{c}$ then it suffices to show $Z_{c}$ is deformable into $X$ to complete the proof.

So first I wanted to know if this is a good way of attacking the proof and if I haven't stated anything incorrectly? If so, then I wanted to know of any way to show $Z_{c}$ is deformable into $X$?

I just started this book and so my knowledge of algebraic topology is only up to the fourth section of Spanier.

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  • $\begingroup$ What is your proof of the first implication? The simplest proof I can think of is easily reversible, so it gives both implications. $\endgroup$ – Eric Wofsey Dec 20 '15 at 1:36
  • $\begingroup$ So I first noted that the inclusion map $i:X \hookrightarrow Z_{f}$ is a cofibration, which is a result from a theorem in Spanier and so it has the ($X$,$Z_{f}$) has the homotopy extension property, so $X$ is only a retract if and only if it is a weak retract. So I only needed to show there is $r:Z_{f}\rightarrow X$ such that $ri \simeq 1_{X}$. We can just take r to be the constant map from $Z_{f}$ to $X$ and so $ri$ is continuous and so since any two maps from an contractible space into a contractible space are homotopic then $ri \simeq 1_{X}$ and thus X is a retract of any cone $Z_{f}$. $\endgroup$ – user1058860 Dec 20 '15 at 1:46
  • $\begingroup$ Oof, that is way more complicated than is necessary! There is a very simple geometrically constructed bijection between contractions of $X$ and retractions from the cone to $X$, as explained in my answer. $\endgroup$ – Eric Wofsey Dec 20 '15 at 1:48
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A space $X$ is contractible iff the map $p:X\to \{*\}$ is a homotopy equivalence, i.e. iff there exists a map $q:\{*\}\to X$ such that $pq\simeq 1$ and $qp\simeq 1$. Since $pq$ is a map from $\{*\}$ to itself, it is always equal to the identity. So the only thing to check is that $qp\simeq 1$. Now the choice of $q$ is equivalent to a choice of a point $a\in X$, and then $qp$ is just the constant function $a:X\to X$.

So $X$ is contractible iff there is a homotopy from the identity map on $X$ to a constant map. Such a homotopy is a map $H:X\times[0,1]\to X$ such that $H(x,0)=x$ and $H(x,1)=a$ for all $x$ and some fixed $a$. By the universal property of quotient spaces, there is a bijection between such maps $H$ and maps $\tilde{H}:X\times[0,1]/X\times\{1\}\to X$ such that $\tilde{H}(x,0)=x$. But $X\times[0,1]/X\times\{1\}$ is just the cone on $X$, and the condition that $\tilde{H}(x,0)=x$ says exactly that $\tilde{H}$ is a retraction from this cone to $X$, its base.

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I think you're trying too hard. Note that for any cone $CX$ over a point $v$, $CX$ deformation retracts onto $v$, and in particular is contractible. By hypothesis $X$ is a retract of a contractible space, so it must be contractible.

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    $\begingroup$ Why is your last statement true? $\endgroup$ – user1058860 Dec 20 '15 at 1:29
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    $\begingroup$ @user1058860 See here. $\endgroup$ – Pedro Tamaroff Dec 20 '15 at 1:41

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