# Extending an ambiently contractible subspace to a contractible subspace.

Let $X$ be a topological space and $U$ a subspace of $X$. $U$ is ambiently contractible if its inclusion mapping into $X$,

$$\iota:U\rightarrow X,$$

is homotopically equivalent to some constant mapping. (One could also define it by saying that $U$ is ambiently contractible if the identity map on $X$, restricted to $U$, is homotopically equivalent to a constant mapping, but in the next paragraph I'm going to start considering $U$ as a space in isolation of $X$ so I prefer the other definition.)

It can be the case that a subspace which is ambiently contractible is not contractible when considered by itself. For example, consider the unit circle $S^1$ as a subspace of $\mathbb{R}^2$ vs by itself.

However, we can extend $U$ to a subspace $W$ of $X$ containing $U$ such that $W$ is contractible in its own right as follows: Let $h$ be a homotopy between $\iota$ and some constant mapping $c$, which takes every point in $U$ to a point $p \in X\setminus U$:

$$h:U\times [0,1] \rightarrow X$$ $$h(u, 0) = \iota(u); \ \ \ h(u,1) = p.$$

Then let $W$ be the union

$$W := \bigcup_{t\in[0,1]} h(U,t)$$

$W$ is then contractible in the subspace topology by construction.

Question: are the the above result and proof correct? I believe that a null-homotopy for $W$ should be similar to a null-homotopy for the cone on $U$.

(EDIT It's apparent to me now after considering the comment that the above construction of $W$ doesn't work; arbitrarily many examples can be given where the construction is followed to the letter but loops are introduced. I'm working on a new approach starting with the cone $CU$ and mapping it into $X$ such that contractibility is preserved.)

(The context of this question is that I'm trying to prove some implications of a space having a low Lusternik-Schnirelmann category, which is defined by means of ambiently contractible open sets, and I'm finding it would be easier to deal with open sets which are contractible in isolation. The ideal would be to simply extend an ambiently contractible open set to a contractible open set, as I've tried to do here.)

• This $W$ need not be contractible. Let $U=\{(1,0)\}$ and $h:[0,1]\to X$ be the map $t\mapsto e^{2\pi i t}$, so $p=(1,0)$. Then $W$ is a circle. Commented Feb 9, 2015 at 18:21
• Good objection; requiring $p \in M\setminus U$ resolves that particular issue.
– Nick
Commented Feb 9, 2015 at 18:28
• The other reason $p \not\in U$ is necessary, which completely slipped my mind, is that if no such $p$ exists then $U$ is the entire space $X$, and then $U$ ambiently contractible implies $X$ contractible trivially.
– Nick
Commented Feb 9, 2015 at 18:38
• I guess simply requiring $p\not\in U$ isn't enough, though, because you could let $p = (-1,0)$ and simply wrap around the circle an extra half-turn so that $W$ is still non-contractible. My construction of $W$ is still too non-restrictive to suffice...
– Nick
Commented Feb 9, 2015 at 18:53

After further consideration, not only is the "proof" given in the question flawed, as demonstrated by the counterexample in the comments, but it's straightforward to construct a space $X$ with an ambiently contractible subset $U$ such that $U$ does not extend to a contractible subspace.

Consider the closed disk, $D^1$. Now let

$$g:S^1\times I \longrightarrow D^1$$

be any homotopy between the inclusion of the circle $S^1$ as the boundary of $D^1$ and any constant mapping

$$c:S^1 \longrightarrow D^1.$$

Construct $W$ as in the original post and one can verify that $W = D^1$.

Now let $q$ be any point in the interior of $D^1$ and $p$ any point on the boundary and let $X = D^1 / \{p,q\}$. By futzing with the identification map $D^1\rightarrow X$ one sees that the image of boundary of $D^1$ in $X$ is still ambiently contractible and that the construction of $W$ as in the original post still yields the entire space $X$. But, the van Kampen theorem can be applied to show that $X$ has nontrivial fundamental group. So, $U$ does not extend to a contractible subspace.

(I hope it's not in bad taste to post an answer to my own question like this. I didn't want to bloat the original post further given that the answer to the question as stated is decidedly negative.)