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

This question arises from problem 8 on pg. 366 of Munkres. Let $X$ be the union of the sets $(1/n) \times I$, $0 \times I$, and $I \times 0$, where $I = [0,1]$, with the topology it inherits from $\mathbb R^2$. I am trying to prove that the inclusion map from the singleton set $x_0 = (0,1)$ to $X$ is a homotopy equivalence, but that $x_0$ is not a deformation retract of $X$.

Intuitively, my idea is as follows: in order to obtain a homotopy from the constant map $x \mapsto x_0$ to the identity map on $X$, we must contract the entire space $X$ down to the horizontal axis, then slide it back up the vertical axis to $x_0$. But, we cannot do so while keeping $x_0$ fixed because (in order to preserve continuity) the horizontal axis must slide down along with the lines $(1/n) \times I$, which are arbitrarily close to it. However, I am having a hard time formalizing this notion. Any suggestions?

share|cite|improve this question
You may also be interested in looking at this (more complicated) variation of the comb space. – t.b. Dec 22 '11 at 17:49
up vote 1 down vote accepted

Arguments in which points cannot be separated by other points by continuity can usually be easily made precise by using sequences (at least when dealing with metric spaces).

Suppose $F: X \times I \to X$ is a retraction, and consider the sequence $x_n = (1/n, 1)$. Since $x_n \to x_0$, by continuity $F(x_n, t_0) \to F(x_0, t_0) = x_0$ for all $t_0$.

Now, take a small enough neighborhood $U$ of $x_0$. By the convergence above, there exist a neighborhood $V_{t_0}$ of $t_0$, and $n_0$ such that, for $t \in V$ and $n \geq n_0$, $F(x_n, t) \in U$. But $U$ can be chosen such that $X_{n} = (\{1/n\} \times I) \cap U$ is clopen in $U$ for all $n$, so $F(x_{n}, t) \in X_{n'}$ for all $t \in V_t$, $n \geq n_0$ and some $n'$.

By using compactness of $[0, 1]$, we can extract a finite subcover from all the $V_t$, and we obtain that for sufficiently large $n$, $F(x_n, t) \in X_{n'}$ for all $t$ including $0$ and $1$, contradiction.

share|cite|improve this answer

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.