Let $X$ be a topological space. I want to show that the cone $CX$ is contractible. Here we construct a deformation retraction from $CX$ to the tip point of the cone $$H_t: CX\to CX;\; (x,t')\mapsto (x,t'(1-t))$$ is this correct?


3 Answers 3


This is an "elementary" exercise, that I've found in lots of lists of problems in a first year of General Topology and in fact shouldn't be asked at that time.

May I ask you how are you going to prove that the map $H$ is continuous?

Chances are that you're not able to do so -but it's not your fault, if this is the first time you're studying Topology.

The reason is that you need some extra resources that, usually, are not taught at this level and are the following:

  • In order to define your map $H$, you are actually first considering a map

$$ H' : (X \times I) \times I \longrightarrow X \times I \ , \qquad H'((x,t),s) = (x, t(1-s)) $$

which is a honest, continuous map, just because of the universal property of the product of spaces. (BTW, you shouldn't "hide" the $I$.)

  • Then you're composing it with the identification map

$$ \pi : X\times I \longrightarrow CX \ , \qquad \pi (x,t) = [x,t] \ , $$

which is fair, since this is also continuous by definition of the quotient topology on $CX$.

  • Next, you can say that this composition $\pi H'$ passes to the "quotient" $CX \times I$ (and here you're entering the risky zone), because it sends all the points you're identifying $(x,0), x \in X$, to the same point, the tip of your cone $[x,0]$:

$$ \pi H' ((x,0),s) = [x,0] \ , \quad \text{for all}\quad x \ . $$

So it induces a well-defined map: your $H$. Right. But: why is this map $H$ continuous?

The reason is that because the natural map $\pi \times \mathrm{id} : (X\times I) \times I \longrightarrow CX \times I$ is an identification; that is, $CX\times I$ has the quotient topology induced by this $\pi \times \mathrm{id}$ and hence, by the universal property of the quotient topology, $H$ is continuous. (Otherwise said: the product topology of $CX \times I$ agrees with the quotient topology induced by $\pi\times\mathrm{id}$.)

But it's not generally true that if you have and identification $\pi : X \longrightarrow Y$, then $\pi \times \mathrm{id} : X\times K \longrightarrow Y\times K$ is necessarily an identification too. This is true when $K$ is a locally compact Hausdorff space, which our $K = I$ fortunately happens to be. So, despite what it looks like, $CX \times I$ is an honest quotient too and everything works. (You can find this result in Bredon's "Topology and Geometry", Springer GTM, proposition 13.19.)

  • $\begingroup$ I am going through your answer and it is very helpful for me so far.Thnx. But I have a question. Universal property of quotient topology says that if $H'$ (here) is constant on fiber of $([(x,t)],s) $then the required map $H$ is continuous. But only ([x,0],s) has fiber non singleton ie it is $\{((x,0),s)\mid x\in X\} $ and I am not sure why $H'$ is constant on this set since the image under $H'$ is $(x,0)$ which is different for different $x$. Sorry, I am not very sound in topology and perhaps missing something. Will you please help me?Thnx in advance. $\endgroup$
    – usermath
    Commented Aug 20, 2015 at 6:28
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    $\begingroup$ Nope. This is not what the universal property of the quotient topology says. What it really says is the following: if $\pi \times \mathrm{id}$ is an identification and $H'$ is continuous and passes to the quotient, then, automatically the induced map $H$ is continuous. So the point here is that you need $\pi \times \mathrm{id}$ to be an identification. $\endgroup$ Commented Aug 21, 2015 at 7:57
  • $\begingroup$ @ a.r Thank you so much. I misunderstood the universal property. The property I mentioned is actually from en.wikipedia.org/wiki/Quotient_space_(topology)#Properties That's why I got confused. However this is clear to me now. Thanks a lot. One request,if you don't mind. Will you please suggest me some article from where I can get an overview of such properties of quotient topology? It will be very helpful for me. Thnx again. $\endgroup$
    – usermath
    Commented Aug 21, 2015 at 13:05
  • 1
    $\begingroup$ You can get one of the best books on general topology for free here: maths.ed.ac.uk/~aar/papers/munkres2.pdf $\endgroup$ Commented Aug 22, 2015 at 14:12
  • $\begingroup$ Not sure if this method works. This is simply because we do not really have an identification map $\pi: X\to Y$ $\endgroup$
    – jk001
    Commented Feb 6, 2022 at 17:06

You have to modify your homotopy a bit to make this work:

Define $H : CX \times I \to CX$ by $H([x,t],s) = [x,(1-s)(t) + t]$ and that should work. Actually on second thoughts your homotopy above should work but your notation is confusing: Define

$$H : CX \times I \to CX$$

by $H([x,t],s) = [x,t(1-s)]$ and that should work, at $s = 0$ you get the identity map on $CX$ and at $s = 1$ you get the constant map at the tip of the cone which for you is $[x,0]$ (my cone tip is $[x,1]$ ) which is different from yours but it does not matter.

  • $\begingroup$ i'm hiding the $I$ to make it simpler but it is the same thing. Thanks! $\endgroup$
    – palio
    Commented Sep 2, 2012 at 12:23
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    $\begingroup$ @fpqc your homotopy does not work because if you take $t=1$ and make $s$ run in $[0, 1]$ then you will get points of the form $[x, t^{'}]$ with $t^{'}> 1$, but those points don't exist. Do you agree? $\endgroup$
    – PtF
    Commented Apr 12, 2014 at 19:14

First define the auxiliary function $f:I\times I\longrightarrow I$ given by $$f(s, t)=(1-s)t+s.$$ Define $H:C(X)\times I\longrightarrow C(X)$ setting $$H([x, t], s)=[x, f(s, t)].$$ It is easy to check $H$ is homotopy from $I_{C(X)}$ to a constant.

Obs: You must take care with the function $f$ to be chosen. You must assure $f$ assumes its values on $I$. For instance, $f(s, t)=2t-s$ wouldn't work because, for example, $f(0, 1)=2$ and there is no such $[(x, 2)]$ on $C(X)$.


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