This is the exercise from Munkres' section 58 Exercise 8.

Find a space $X$ and a point $x_{0}$ of $X$ such that the inclusion $\{x_{0}\}\hookrightarrow X$ is a homotopy equivalence but $\{x_{0}\}$ is not a deformation retract of $X$.

Munkres gives some hint as follows:

Let $X$ be the subspace of $\mathbb{R}^{2}$ that is the union of the line segments $(1/n)\times [0,1]$, for $n\in\mathbb{Z}_{+}$, the line segment $0\times [0,1]$, and the line segment $[0,1]\times 0$; let $x_{0}$ be the point $(0,1)$. If $\{x_{0}\}$ is a deformation retract of $X$, then show that for any neighborhood $U$ of $x_{0}$, the path component of $U$ containing $x_{0}$ must contains a neighborhood of $x_{0}$.

Then I tried to follow the hint to do this exercise. However, it is even hard to prove the inclusion $i:\{x_{0}\}\hookrightarrow X$ is a homotopy equivalence.

Let $r:X\longrightarrow \{x_{0}\}$ be the retraction of $X$ onto $\{x_{0}\}$ be defining it as $r(x):=x_{0}$ for all $x\in X$. Then, it is clear that $r\circ i=Id_{\{x_{0}\}}$.

However, without the convexity, it is hard to show $i\circ r\sim Id_{X}$. Thus the problem reduces to how to show such $X$ is convex?

Now, suppose by contradiction that $\{x_{0}\}$ is a deformation retract of $X$, then there is a continuous map $F:X\times [0,1]\longrightarrow X$ such that $H(x,0)=x$ and $H(x,1)=x_{0}$ for all $x\in X$, and $H(x_{0},t)=x_{0}$ for all $t\in[0,1]$.

Then I really don't know how to proceed..

Also, even thought I showed the hint, where would be the contradiction?

I may need an answer with details since I am not really familiar with path-component.

Thank you!


1 Answer 1


First, $X$ is clearly not convex. That's obvious from a picture, and it's also needed since any point in a convex set is a deformation retract of the set.

The key difference between "homotopy equivalence" (which you want to be true) and "deformation retract" (which you want to be false) is that for the former you need a homotopy from the identity map of $X$ to $i\circ r$ (the constant map sending all of $X$ to the point $x_0$), whereas for the latter you need such a homotopy that never moves $x_0$.

The former is pretty easy, despite the lack of convexity. First deform all points of $X$ straight down into $[0,1]\times0$; then deform that horizontally to the point $(0,0)$; finally, slide that point up to $x_0$.

The main thing you need to prove is that you can't get such a homotopy without moving $x_0$. The key ingredient of that proof will be that, because $x_0$ is unmoved, very nearby points $(\frac1n,1)$ can't move very far, therefore can't get down to $[0,1]\times 0$, and therefore can never get to $x_0$ (or to any point on the $y$-axis).

  • $\begingroup$ Thank you so much! $\endgroup$ Nov 10, 2019 at 19:58
  • $\begingroup$ Is that possible for you to write down the exact equation of the homotopy for the homotopy equivalence and the deformation retract? $\endgroup$ Nov 10, 2019 at 21:43
  • $\begingroup$ To prove the key ingredient, we assume such a homotopy $H: X \times I \rightarrow X$ exists. For any neighborhood $U$ of $x_0$, we have $x_0 \times I \subset H^{-1}(U)$, where $H^{-1}(U)$ is open. Now notice that I = [0,1] is compact and use Lemma 26.8 (the tube lemma) in Munkres. $\endgroup$ Nov 25, 2023 at 22:54

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