Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I already asked about the interpretation of this problem here. Now I would like to ask about the solution. The problem is

Let $A\subseteq X$ be a contractible space. Let $a_0\in A$. Is the embedding $f: X\setminus A\to X\setminus\{a_0\}$ a homotopy equivalence?

I'm not very bright, and this is very difficult for me. I think I've solved this problem though:

Let $X=[0,1]\cup\{2\},\, A=[0,1],\,a_0=0.$ Then $X\setminus A=\{2\}$ and $X\setminus\{a_0\}=(0,1]\cup\{2\}.$ There is exactly one function $g:(0,1]\cup\{2\}\to \{2\}$ so I only have to check if $f\circ g$ and $g\circ f$ are homotopic to the appropriate identity functions. $g\circ f=\mathrm{id}_{\{2\}}$, so this one is OK. I have to check if $f\circ g$ is homotopic to $\mathrm{id}_{(0,1]\cup\{2\}}.$ Suppose there is a homotopy $H(x,t).$ Then $H(1,t)$ is a path from $1$ to $2$ in $(0,1]\cup \{2\},$ which is a contradiction.

So it's not true, if that is correct. But I took a non-connected $X$. I remember that during the course some verbal agreements were made that we generally consider all spaces connected and Hausdorff, and maybe even more. Unfortunately, I don't remember it very well. And I'm not sure the agreement applied to this problem. But regardless of that I would like to know if this is still false when $X$ is connected. Or even more: what conditions on $X$ do I need for this to be true?

I'm asking this because I feel that I'm cheating in my solution. I feel I haven't really understood the problem, despite a lot of time and effort I've put into it.

share|cite|improve this question
up vote 1 down vote accepted

I think your argument is correct, indeed since path-connectedness is invariant under homotopy equivalence the two spaces $X \setminus A$ and $A \setminus a_0$ cannot be homotopy equivalent - the first is path-connected while the second is not.

A similar example when $X$ is connected is $X = [0,2]$, $a_0 = 1$ and $A = [0,1]$. Then $X \setminus A = (1,2]$, and $X \setminus a_0 = [0,1) \cup (1, 2]$ which are not homotopy equivalent, so in particular the embedding cannot be a homotopy equivalence.

share|cite|improve this answer
Thanks, I understand. So I think it doesn't really make sense to ask about conditions on $X$. It's also both $A$ and $a_0$ that matter. – Bartek Aug 22 '12 at 17:50

Please do not say that you are not bright. You have done some great mathematics here.

The counterexample you have given is almost correct. The last line should be "Then $H(1,t)$ is a path from $(1,0)$ to $(2,1)$ in $(0,1] \cup \lbrace 2\rbrace \times [0,1]$, which is a contradiction.

You have not cheated the solution at all, but have done what any good mathematician should always do: figure out how it could be false and what conditions might make it true.

Now let us suppose that the space $X$ is connected and we again have $A \subseteq X$ where $A$ is contractible and $a_0 \in A$. For the sake of simplicity, let us consider $X=[0,1]$. Can you think of a way to choose $A$ to make the embedding $f$ not be a homotopy equivalence? It is ok if you feel like your choice of $A$ is "cheating" again.

share|cite|improve this answer
Thank you. It seems that Simon StJohn-Green has already given the solution. I see that it is still false now. – Bartek Aug 22 '12 at 17:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.