I'm trying to solve a problem from Hatchers "Algebraic Topology" - exercise 0.6 (a):

Let $X$ be the subspace of $\mathbb{R}^{2}$ consisting of horizontal segment $[0,1]\times \lbrace 0 \rbrace$ together with all the vertical segments $\lbrace r \rbrace \times [0,1-r]$ for $r$ rational number in $[0,1]$. Show that $X$ deformation retracts to any point in the segment $[0,1]\times \lbrace 0 \rbrace$, but not to any other point.

In the exercise there is a hint "See the preceding problem". In the preceding problem I have obtained the result:

"If a space $X$ deformation retracts to a point $x$, then for each neighborhood $U$ of $x$ in $X$ there exists a neighborhood $V\subset U$ of $x$ such that the inclusion $V\hookrightarrow U$ is nullhomotopic"

When it comes to solving these types of exercises I have some serious problems. In this exercise I have problems with the part: "but not to any other point". I might be lost (completely in the theory) but here is my attempt to show that "any other point is not a deformation retract of $X$":

My attempt:

First I assume for contradiction that $x_{0} \in \lbrace r \rbrace \times [0,1-r]$ is a deformation retract of $X$. I now use the preceding exercise. I choose my $U$ to be a set consisting of infinite many segments (not just one):

From the preceding exercise I know there exists a $V\subset U$ of $x_{0}$ such that the inclusion map $V\hookrightarrow U$ is null homotopic. That is there exists a homotopy $F: V\times I \rightarrow U$ such that:

  • $F(x,0) = x$
  • $F(x,1) = c$ where $c$ is a constant in $U$.

I can extend this homotopy on $U$ since $(U,V)$ has the homotopy extension property (?) and from this I conclude that the new homotopy $F: U\times I \rightarrow U$ gives us that $U$ is contractible and this is a contradiction (if we look at the picture).

My question(s):

  • Does this argumentation makes sense? And how do I know that $(U,V)$ has the homotopy extension property (I have tried to deduce that it was a $CW$ pair, but that requires $V$ to be closed and contractible). And if not, how do I then show it?

Instead of just "defining" $U$ as "consisting of infinite many segments", you should define it explicitly -- for example as all of $X$ except for the horizontal base. (Actually, since $U$ has to be a neighborhood of $x_0$, it has to contain (parts of) infinitely many teeth of the comb, so your choice is not a choice at all).

The last step is a little roundabout. It would be simpler to note that any neighborhood $V\subseteq U$ of $x_0$ contains points with a different first coordinate, and such a point has no path to $x_0$ in $U$.


Hint: If $f\colon Y\rightarrow X$ is a nullhomotopic map, then $f(Y)$ is a subset of some $U\subset X$ such that $U$ is a connected subset of $X$.

  • 1
    $\begingroup$ I think you are wrong here. Take $F:\left\{ 0,1\right\} \times\mathbb{I}\rightarrow\mathbb{R}$ with $\left(0,t\right)\mapsto t$ and $\left(1,t\right)\mapsto1$. Then $F:f\simeq c$ where $f:\left\{ 0,1\right\} \rightarrow\mathbb{R}$ is the inclusion and $c:\left\{ 0,1\right\} \rightarrow\mathbb{R}$ is the map constant at $1$. Then $f$ is nullhomotopic, but its image $\left\{ 0,1\right\} $ is not a connected subset of $\mathbb{R}$. A correct thing to say of a null-homotopic $f:Y\rightarrow X$ is that $f\left(Y\right)$ meets only one path-component of $X$. $\endgroup$
    – drhab
    Sep 24 '13 at 15:54
  • $\begingroup$ @drhab Sorry yes you're right. $\endgroup$
    – Dan Rust
    Sep 24 '13 at 16:08

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