A complete exercise of Algebraic topology [closed]

Question

My teacher of Algebraic topology has shared exams from past years and this exercise appears on them:

Consider the following sets on the Euclidean plane $\Bbb{R}^2$: $$X_1 = \{(x,y) \in \Bbb{R}^2 : x^2+y^2=1 \} \\ X_2 = \{(x,y) \in \Bbb{R}^2 : 1 \leq x \leq 2 \, ; y=0 \} \\ X_3 = \{(x,y) \in \Bbb{R}^2 : (x-3)^2 + y^2 \leq 1\} \\ X = \bigcup_{k=1}^3 X_k$$

1. Compute, reasonably, the fundamental group of $X$.
2. For each $n\in \Bbb{Z}, n\geq 0$ compute a compact, path connected subspace $K_n \subseteq X$ whose fundamental group is the free group of degree $n$.
3. Describe a simply connected covering space $(\tilde{X}, p)$ of $X$ and describe clearly the $p$.
4. Let denote $Y=S^2 \vee \Bbb{R} \Bbb{P}^2$ and let $f: Y \rightarrow X$ a continuous map. Prove that $f$ is null-homotopic.
5. Let $k \in \Bbb{Z}, k \geq 1$. Construct a connected covering space $(\tilde{X_k}, p_k)$ of $X$ with exactly $k$ leafs. Let $x_0 \in X$ and for all $k \geq 1$ take $\tilde{x}_0^k \in p_k^{-1}(x_0)$. Determine the subgroup $H_k = {p_k}_\ast (\pi_1 (\tilde{X_k}, \tilde{x}_0^k) )$ of $\pi_ 1(X, x_0)$.

I am trying to solve it, so I want to know if my ideas are fine and what to do to complete it.

For 1., I think that, since the space $X$ is $\Bbb{S}^1$ $(X_1)$ connected by a segment $(X_2)$ with $\Bbb{D}^2$ $(X_3)$, it is clear that $\Bbb{S}^1$ is deformation retract of $X$, so $\pi_1 (X)= \pi_1 (\Bbb{S}^1) = \Bbb{Z}$.

For 2., I do not have clear what to do. It cames to me ideas like set $n-1$ circles indise of $X_3$ pairwise connected by a single point and one of them connected to $X_2$ on $(2,0)$, but I am not sure about this.

For 3., I am a little confused since my intuition says that, since $\Bbb{S}^1$ is deformation retract of $X$, it has to be some way to show that $\Bbb{R}$, who is the universal covering space of $\Bbb{S}^1$, it is the space looked for.

For 4., I have seen a theoretical result that every continuous map $f: X \rightarrow S^1$ is null-homotopic iff the induced homomorphism is trivial. Therefore, I think that it can be proved as well showing that the given map $f$ induces $f_\ast : \pi_1(Y) = \Bbb{Z}_2 \rightarrow \Bbb{Z}$, but I do not know how to do this.

For 5., I am very stuck and do not know how to solve it, to be honest.

Thanks in advance :)

Answer 1

For 1., I think that, since the space $X$ is $\Bbb{S}^1$ $(X_1)$ connected by a segment $(X_2)$ with $\Bbb{D}^2$ $(X_3)$, it is clear that $\Bbb{S}^1$ is deformation retract of $X$, so $\pi_1 (X)= \pi_1 (\Bbb{S}^1) = \Bbb{Z}$.

Looks correct.

For 2., I do not have clear what to do. It cames to me ideas like set $n-1$ circles indise of $X_3$ pairwise connected by a single point and one of them connected to $X_2$ on $(2,0)$, but I am not sure about this.

Indeed. Except $n$ circles. The formula would be something like this:

$$C_i=\{(x,y)\ |\ (x+1/i)^2+y^2=(1/i)^2\}$$ $$K_n=\bigcup_{i=1}^n C_i$$

Then $K_n$ topologically is homeomorphic to $\bigvee_{i=1}^n S^1$ which is well known to have free fundamental group of rank $n$.

For 3., I am a little confused since my intuition says that, since $\Bbb{S}^1$ is deformation retract of $X$, it has to be some way to show that $\Bbb{R}$, who is the universal covering space of $\Bbb{S}^1$, it is the space looked for.

No. Universal covering is uniquely determined up to homeomorphism, not homotopy equivalence. For example every contractible space is universal covering of itself, but clearly these don't have to be homeomorphic.

In this particular case $\Bbb{R}$ isn't even a covering of $X$, because every covering map is a local homeomorphism. Clearly the two-dimensional piece cannot be locally homeomorphic with $\Bbb{R}$.

Generally universal covering of wedge sum of two spaces is complicated. But it is quite straightforward if one of the pieces is simply connected. First take $\Bbb{R}$ (which is universal covering of $S^1$) and then glue to it a copy of $X_2\vee X_3$ at every integer $n\in\mathbb{Z}$. The resulting space is contractible and there's an obvious covering onto $X$, and thus it is the universal covering. The drawing:

For 4., I have seen a theoretical result that every continuous map $f: X \rightarrow S^1$ is null-homotopic iff the induced homomorphism is trivial. Therefore, I think that it can be proved as well showing that the given map $f$ induces $f_\ast : \pi_1(Y) = \Bbb{Z}_2 \rightarrow \Bbb{Z}$, but I do not know how to do this.

You are almost there. You literally have to notice that there is no non-trivial group homomorphism $\mathbb{Z}_2\to\mathbb{Z}$ because $\mathbb{Z}$ has no non-trivial finite subgroups.

For 5., I am very stuck and do not know how to solve it, to be honest.

This is similar to the idea for the universal covering. First you take a covering $p_k:S^1\to S^1$ with $k$ leafs, i.e. $p_k(z)=z^k$ in terms of complex numbers. Then you attach copy of $X_2\vee X_3$ at every $k$-th root of unity in $S^1$. It is tempting to do the attaching in a smart way, i.e. by copying $X_2\vee X_3$ and rotating $k$ time. With such construction simple rotation becomes the covering map we are looking for. However the devil is in the details: for big enough $k$ copies of $X_2\vee X_3$ will intersect each other. And we don't want that. Therefore we need to do an abstract glueing at each $k$-th root of unity, and then consider a "projection" as the covering map. Here we project only copies of $X_2\vee X_3$ onto $X_2\vee X_3$, the base $S^1\to S^1$ stays as rotation $z\mapsto z^k$. The drawing for $k=4$:

As for the subgroup this requires some tedious calculations which follow from general properties of coverings. Both spaces have $\mathbb{Z}$ as the fundamental group and it can be shown that the induced morphism $(p_k)_*$ maps $[x]$ to $k[x]$. And so the subgroup we are looking for is $k\mathbb{Z}\subseteq\mathbb{Z}$. I'm not going to go into details on this one, feel free to ask a separate question.