# A homotopy question

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.

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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.

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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.

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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