Is Armstrong saying that the comb space is not contractible?

Question

That seems to be what it says on page 108 of Basic Topology by M. A. Armstrong. There, it says

In 'homotoping' the identity map $1_X$ to a constant map $c_p$, we may be forced to move the point p during the homotopy, i.e., there may not be a homotopy relative to $\{p\}$ from $1_X$ to $c_p$. For example, take the 'comb space' shown in Fig. 5.10 as X, and take p to be the point (0, 1/2). There is no homotopy from $1_X$ to $c_p$ which keeps p fixed. (Why not?)

"Why not?" indeed! I do not understand why he says we might be forced to move p.

For one thing, the definition of contractible does not mention that the homotopy from the identity function to $c_p$ needs to be relative to $\{p\}$, but for the sake of argument, let us keep that constraint.

Moreover, it seems to me that you can construct a homotopy that is relative to $\{p\}$ in aforementioned case involving the comb space as follows:

1. Collapse all of the teeth, except the left most one down to the "spine" (i.e. $I \times \{0\}$). This leaves you with an L shape.
2. You can then collapse the "spine" of the comb to the bottom of the one remaining tooth. Then, you are left with a single line segment containing the point that you want to collapse down to.
3. This space is convex; therefore, you can collapse down to p by straight lines.

Am I misunderstanding that section of text? Or does it not make sense? Or is there something else that's going wrong?

PS: It seems odd to me that Armstrong defines the comb space as having height 1/2 instead of 1, which seems to be how various sources on the Web do it, but I suppose that makes no difference.

Answer 1

The comb space is contractible. Collapse all of the teeth including the left most one down to the spine. Then collapse the spine to one of its points.

By the way, the height of the comb space is irrelevant, variants with different heights are homeomorphic.

A space $X$ being contractible means that the identity map on $X$ is homotopic to a constant map, i.e. that there is a homotopy $H : X \times I \to X$ such that for all $x$ we have $H(x,0) = x$ and $H(x,1) = c$ for some $c \in X$. In that case we say that $X$ is contractible to the point $c$.

If $X$ is contractible, then it is contractible to any point $d \in X$. To see this, define $$G : X \times I \to X, G(x,t) = \begin{cases} H(x,2t) & t \le 1/2 \\ H(d,2(1-t)) & t \ge 1/2 \end{cases} $$

This is well-defined since for $t = 1/2$ we have $H(x,2t) = H(x,1) = c = H(d,1) = H(d, 2(t-1/2))$. The map $G$ is a homotopy from the identity to the constant map with value $d$.

A stronger concept is pointed contractibility. We say that $X$ is pointed contractible to $c \in X$ if the above homotopy $H$ has the additional property that $H(c,t) = c$ for all $t$. This means that the point $c$ is not moved during the contraction. A contractible space need not be pointed contractible to an arbitrary point.

Armstrong gives an example for this phenomenon. In fact, the comb space is not pointed contractible to $p$ (defined to be $(0,1/2)$). Assume that there is a contraction $H$ keeping $p$ fixed. The set $U = X \cap I \times (0,1]$ is an open neigborhood of $p$ such that $H(\{p\} \times I) \subset U$. Thus $\{p\} \times I) \subset V = H^{-1}(U)$. Since $V$ is open in $X \times I$, there exists an open neigborhood $W$ of $p$ in $X$ such that $W \times I \subset V$. This is true because $I$ is compact. Thus $H(W \times I) \subset U$. For each $x \in W$ the set $H(\{x\} \times I)$ is path connected and contains $p = H(x,1)$. Thus $x = H(x,0)$ and $p = H(x,1)$ must be contained in the same path component of $U$. But this cannot be true because $W$ contains the points $x_n = (1/2,1/n)$ for $n \ge n_0$. Note that $p$ and $x_n$ belong to different path components of $U$.

Answer 2

For one thing, the definition of contractible does not mention that the homotopy from the identity function to $c_p$ needs to be relative to $\{p\}$

Indeed, it doesn't have to be. But Armstrong doesn't claim that the comb space is not contractible. He claims that every such homotopy has to move points from the first vertical line. In other words he says that homotopy and relative (or pointed) homotopy are not the same concepts.

1. Collapse all of the teeth, except the left most one down to the "spine" (i.e. $I \times \{0\}$). This leaves you with an L shape.

Your construction already fails at the first step: this operation is not continuous. Of course under the assumption that the $(0,1/2)$ point is stationary during the collapse. Indeed, consider the sequence $(1/n,1/2)$ with its (stationary) limit $(0,1/2)$ and think if a homotopy $H$ can actually do what you want in a continuous way.

PS: It seems odd to me that Armstrong defines the comb space as having height 1/2 instead of 1, which seems to be how various sources on the Web do it, but I suppose that makes no difference.

No, he only took $p$ to be $(0,1/2)$. I think you overinterpret this. The height can be arbitrary, the choice of $p$ is quite arbitrary as well, it works for any $p=(0,t)$ and $t>0$. I suppose that the choice of $1/2$ is a bit misleading. But only a bit.