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.

Thank you in advance!

  • 1
    $\begingroup$ 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 $\endgroup$ Aug 6, 2019 at 11:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paul Frost
    Aug 6, 2019 at 21:05
  • $\begingroup$ @PaulFrost oh my, you're right of course $\endgroup$ Aug 6, 2019 at 21:06

1 Answer 1


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 $Z$.


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.

  • $\begingroup$ Thank you so much for your answer! You've been awarded! $\endgroup$
    – Albert
    Aug 10, 2019 at 21:30

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