Computing the fundamental group of a union of subspaces

Question

If $x, y \in \Bbb{R}^n$, we denote by $[x,y] := \{ (1-t) \, x + t \, y : t \in [0,1] \subset \Bbb{R} \}$

Let denote the following subspaces of $\Bbb{R}^2$:

$$ X_1 := \{ x \in \Bbb{R}^2 : ||x||=1 \} \\ X_2 := [(1,0),(2,0)] ; \, X_3 := [(2,0), (3,1)] ; \, X_4 := [(2,0), (3,-1)] \\ X:= \cup_{k=1}^4 X_k$$

As an excercise of my Algebraic topology lessons, I have to compute the fundamental group of $X$.

I intuit that it has to be $\Bbb{Z}$ since $X_3$ and $X_4$ are contractible to $\{(2,0)\}$, $X_2$ is contractible to $\{(1,0)\}$ and then $X$ is contractible to $X_1 = \Bbb{S}^1$ whose fundamental group is $\Bbb{Z}$, but I have not much experience computing fundamental groups so confirmation or corrections would be very appreciated :)

$X_1, X_2, X_3$ and $X_4$ are respectively red, purple, black and blue" />

Answer 1

Yes, the fundamental group of $ X:= \bigcup_{k=1}^4 X_k$ is isomorphic to $\mathbb{Z}$, since $X_1$ is a deformation retract of $X$. It's a bit tiresome to write out the formula, but clearly the retraction shrinks $X_2\cup X_3\cup X_4$ to $(1,0)$ and keeps the circle intact.

There is another way to think about this. $X_2\cup X_3\cup X_4$ is a wedge of the spaces $X_2, X_3, X_4$, thus its fundamental group is the external free product of the fundamental groups of the spaces $X_2, X_3, X_4$, all of which are trivial. And again $X$ is a wedge of $X_1$ and $X_2\cup X_3\cup X_4$, so $\pi_1(X,x_0)=\mathbb{Z}*\{e\}=\mathbb{Z}$.