# Is this comb-like space contractible?

For each $$\ n \in \mathbb{N}^* = \{ 1,2,3,4,... \}$$, let $$\ S_n = \big\{ (t,1-nt) \in \mathbb{R}^2 : 0 \leqslant t \leqslant 1/n \big\}$$, $$Y_n = \big\{ (t,-nt-1) \in \mathbb{R}^2 : -1/n \leqslant t \leqslant 0 \big\}$$, $$\displaystyle S = \bigcup_{n \in \mathbb{N}^*} S_n$$, $$\displaystyle Y = \bigcup_{n \in \mathbb{N}^*} Y_n \$$ and $$\ Z = \{ 0 \} \times [-1,1]$$. Consider $$\ X = Y \cup Z \cup S$$, with the subspace topology inherited from the euclidean usual topology of $$\, \mathbb{R}^2$$. I have attached a picture of the space below.

My questions are

$$(1) \$$ Is this space $$X$$ contractible? Why? How can I prove it?

$$(2) \$$ How can I compute the homotopy groups of $$X$$?

I think the answer for $$(1)$$ is no. I have tried to prove it by contradiction. Suppose it is. So, we have homotopy equivalences $$\ f : X \to * \$$ and $$\ g: * \to X \$$ such that $$\ f \circ g \sim id_* \$$ and $$\ g \circ f \sim id_X \,$$. Then, there exists a homotopy $$\ H: X \times I \to X \$$ such that $$\ H(x,0) = x \$$ and $$\ H(x,1) = g \big( f(x) \big),$$ $$\forall x \in X$$. I think this will let me to some contradiction, but I am stuck.

For $$(2)$$ it is even worse. I have tried to visualize the image of some general continuous map $$\ f : S^n \to X \$$ in $$X$$, but I see nothing.

Any help will be highly appreciable.

• your first question is (almost) a duplicate of math.stackexchange.com/questions/423640/… I think the answers there apply here just fine Commented Apr 2, 2016 at 11:42
• @Najib - As far as I can tell we're not including the interval $[-1,1]$ on the x-axis, so it looks like the union of two contractible combs as opposed to two very non contractible Hawaiian earrings to me.
– user98602
Commented Apr 2, 2016 at 12:05
• @Mike Oh you're absolutely right. I misread the question (more precisely the $Z$, I swapped the variables in my head). Commented Apr 2, 2016 at 12:24

Claim: Any continuous map $f:S^n\to X$ can be homotoped to a map whose image is contained in $S\cup Z$.
Given $f:S^n\to X$, define a homotopy
$$\begin{array} &H:S^n\times I\to X&\\ (x,t)\mapsto\begin{cases} f(x) &\mbox{if } f(x)\in S\cup Z \\ tf(x)+(1-t)(0,-1) & \mbox{if } f(x)\in Y \end{cases} \end{array}$$ $H$ is well-defined, continuous, $H_1=f$ and the image of $H_0$ lies in $S\cup Z$.
This proves that $\pi_n(X)=0$ since $S\cup Z$ is contractible (it deformation retracts onto $(0,1)$).