# Homotopy groups of double comb space

Let $Y$ be the comb space, that is the following subspace of $\mathbb{R}^2$: $$Y = (I\times\{0\})\cup (\{0\}\times I) \cup \bigcup_{n\in\mathbb{N}^*} (\{1/n\}\times I),$$ where $I=[0,1]$ and $\mathbb{N}^*=\mathbb{N}\setminus\{0\}$. Let $y_0=(0,1)$ and let $Y'$ be another copy of $Y$ with corresponding point $y_0'$. Let $X$ be the wedge sum of $Y$ and $Y'$ obtained by identifying the points $y_0$ and $y_0'$. The space $X$ is called the double comb space.

Note that $X$ is clearly homeomorphic to the following subspace of $\mathbb{R}^2$: $$(\{0\}\times [-1,1]) \cup ([0,1]\times\{1\})\cup ([-1,0]\times\{-1\})\cup \bigcup_{n\in\mathbb{N}^*} ((\{1/n\}\times [0,1]) \cup (\{-1/n\}\times [-1,0]))$$

It is clear that $X$ is a path connected space and I have proved that $X$ is non contractible. I need to prove that the homotopy groups $\pi_n(X)$ are trivial for every $n\in\mathbb{N}^*$ by proving that every map $f:S^n\to X$ is homotopic to a constant map (or, equivalently, by showing that every such map can be continuously extended over the $n+1$ dimensional disc).

There is a similar problem with the double comb-like space, that is, the following subspace of $\mathbb{R^2}$: $$Z = \{0\}\times [-1,1]\cup \bigcup_{n\in\mathbb{N}^*} ([(-1/n,0),(0,-1)]\cup [(1/n,0),(0,1)])$$ where $[a,b]$ is the line segment joining the points $a$ and $b$ of $\mathbb{R}^2$. Here I have proved that $\pi_n(Z)=0$ for every $n$ by showing that every map $S^n\to Z$ is homotopic to a constant map.

I have tried to adapt the construction of the homotopy in the case of the space $Z$ to the case of the space $X$, but I failed. Also I tried to study if the spaces $X$ and $Z$ are of the same homotopy type, but I have failed again.

I am asking for a direct proof that every map $S^n\to X$ is homotopic to a constant map, if it is possible.

• Out of compactness, you can argue that the image of $S^n \to X$ only meets finitely many $1/n$-branches of the comb, and then it lands in a contractible subspace, and is therefore nullhomotopic Aug 6, 2019 at 11:38
• @Max It can meet all $1/n$-branches of the comb, but the "penetration depth" goes to $0$ as $n \to \infty$. See my answer. Aug 6, 2019 at 21:05
• @PaulFrost oh my, you're right of course Aug 6, 2019 at 21:06

That the homotopy groups of $$Z$$ are trivial has been proved in the answer to Is this comb-like space contractible? (although in my opinion the continuity of the introduced homotopy should be properly proved).

The space $$Z$$ is obtained from $$X$$ by collapsing the two line segments $$[0,1] \times \{1\}$$ and $$[-1,0] \times \{-1\}$$ to points. Using this it can be shwon that the spaces $$X$$ and $$Z$$ are homotopy equivalent. I do not know a completely elementary proof but would have to invoke some facts about cofibrations.

So let us do it without using that $$X \simeq Z$$.

Consider any map $$f : S^m \to X$$. We shall show that $$f(S^m)$$ is contained in a contractible subset $$X'' \subset X$$ which shows that $$f$$ is inessential.

Define $$A_n = f(S^n) \cap \{1/n\}\times [0,1]$$ and $$a_n = \sup \{{t\in [0,1]} \mid A_n \subset \{1/n\} \times [t,1] \}$$. If $$a_n < 1$$, then $$A_n$$ must be nonempty and we see that $$(1/n,a_n) \in A_n$$ because $$A_n$$ is compact. We claim that $$a_n \to 1$$. Suppose this is false. Then $$(a_n)$$ must have a cluster point $$a < 1$$. Let $$(a_{n_k})$$ be a subsequence converging to $$a$$ such that all $$a_{n_k} < 1$$. We have $$(1/n_k,a_{n_k}) = f(x_k)$$ for some $$x_k \in S^m$$. $$(x_k)$$ has a convergent subsequence $$(x_{k_r})$$ with limit $$\xi \in S^m$$. We conclude $$f(\xi) = \lim_r f(x_{k_r}) = \lim_r (1/n_{k_r},a_{n_{k_r}}) = (0,a) \in V = X \cap [-1,1] \times [-1,1)$$. Let $$U$$ be a connected neighborhood of $$\xi$$ (e.g. $$U = S^m \cap B$$ with a some open ball $$B \subset \mathbb R^{m+1}$$) such that $$f(U) \subset V$$. There exists $$k$$ such that $$x_k \in U$$. Then both $$f(\xi) = (0,a)$$ and $$f(x_k) = (1/n_k,a_{n_k})$$ are contained in the connected subset $$f(U)$$ of $$V$$. This is a contradiction because the two points belong to distinct components of $$V$$.

Hence $$f(S^m) \subset X'' = X' \cup \bigcup_{n\in\mathbb{N}^*} \{1/n\}\times [a_n,1]$$ with $$X' = \{0\}\times [-1,1] \cup [-1,0]\times\{-1\} \cup \bigcup_{n\in\mathbb{N}^*} \{-1/n\}\times [-1,0] \cup [0,1]\times\{1\}$$. We claim that $$X'$$ is a strong deformation retract of $$X''$$. But obviously $$X'$$ is contractible (it contains $$[-1,0]\times\{-1\}$$ a strong deformation retract) which finishes the proof.

Define $$H : X'' \times [0,1] \to X'', H(x,s,t) = \begin{cases} (x,t) & (x,t) \in X'\\ (x,s + (1-s)t) & (x,t) \in \bigcup_{n\in\mathbb{N}^*} \{1/n\}\times [a_n,1] \end{cases} \quad.$$ We have $$H(x,t,0) =(x,t), H(x,t,1) \subset X'$$ for all $$(x,t)$$ and $$H(x,t,s) =(x,t)$$ for all $$(x,t) \in X'$$ and $$s \in [0,1]$$. It remains to show that $$H$$ is continuous. This is completely obvious in all points $$(x_0,t_0,s_0)$$ such that $$(x_0,t_0)$$ is contained in the open set $$X'' \setminus (\{0\} \times [0,1])$$. If $$x_0 = 0$$ and $$t_0 \in [0,1)$$, then choose $$r \in (t_0,1)$$. There exists $$N$$ such that $$a_n > r$$ for $$n \ge N$$. Hence $$(-1,1/N) \times (-1,r) \cap X''$$ is an open neighborhood of $$(0,t_0)$$ in $$X''$$ which does not contain any points of $$\bigcup_{n\in\mathbb{N}^*} \{1/n\}\times [a_n,1]$$. This shows continuity in this case. Finally we consider the point $$(0,1)$$. Let $$V$$ be an open neigborhood of $$H(0,1,s_0) =(0,1)$$ in $$X''$$. We may assume $$V = (r,1] \times [0,\varepsilon) \cap X''$$ for suitable $$r, \varepsilon$$. Then for $$(x,t,s) \in V \times I$$ we get $$H(x,t,s) \in V$$.

Note that this proof can easily be adapted for the space $$Z$$.

Edited:

I realized that the proof can easily be generalized to show that any map $$f : Y \to X$$ defined on a sequentially compact locally connected space $$Y$$ is inessential.

Local connectedness is an indispensable condition. In fact, $$X$$ is not locally connected and the identity on $$X$$ is not inessential.

• Thank you so much for your answer! You've been awarded! Aug 10, 2019 at 21:30