fundamental group of $S^{1}\cup\left(\mathbb{R}\times0\right)$ (Theta space) I was asked to find the fundamental group of $S^{1}\cup\left(\mathbb{R}\times0\right)$.
I was given three options - $\mathbb{Z}$, the fundamental group of two spheres connected by one point (i.e F2 - Free Group on 2 Generators) or the trivial group.
I know it's enough to calculate the fundamental group for  $S^{1}\cup\left([-1,1]\times0\right)$.
I tried finding some f and g to show a homotopy equivalency to the circle or to the two connected circles, but failed.
after that, I tried to find a space and a cover map to show that the group is abelian (and therefor isomorphic to Z) or non abelian (and therefor isomorphic to F2).
looking for help and some intuition for the topic.
I think it should be isomorphic to F2, but I don't know how to show it.
one example for something I tried (to show homotopy equivalency to S1) is:
$$f:S^{1}\rightarrow S^{1}\cup\left(\left[0,1\right]\times0\right),\quad f\left(\left(x,y\right)\right)=\left(x,y\right) $$
$$g:S^{1}\cup\left(\left[0,1\right]\times0\right)\rightarrow S^{1},\;g\left(\left(x,y\right)\right)=\frac{\left(x,y\right)}{\|\left(x,y\right)\|} $$
than:
$$ g\circ f:S^{1}\rightarrow S^{1},\quad g\circ f\left(\left(x,y\right)\right)=\left(x,y\right) $$
but I have a problem cause I might divide by 0.
EDIT: This space is known as Theta space.
 A: You can contract the line to the interval -diameter of the circle. So you space is homotopic to a circle crossed by an interval. Further contracting the interval to a point. you get two circles connected by a point. So the fundamental group is free of rank 2.
A: I assume that by this union you mean a disjoint union.
First of all, let's think about a loop in $X$. This is a continuous function from an interval $I$ to $X$. Since $I$ is connected, the image has to be connected. So a loop lies entirely on $S^1$ or on $\mathbb{R}\times\{0\}$. If our base point $x_0$ is a point of $S^1$, then the only loops based on $x_0$ are loops that lie on $S^1$ and are based on $x_0$. Hence in that case it is $\pi_1(X,x_0)=\pi_1(S^1)=\mathbb{Z}$. Otherwise, if $x_0$ is a point of $\mathbb{R}\times\{0\}$, the only loops based at $x_0$ lie at $\mathbb{R}\times\{0\}$. So, $\pi_1(X,x_0)=\pi_1(\mathbb{R}\times\{0\})=\pi_1(\mathbb{R})=$trivial.
EDIT: On the other hand, if you mean that $S^1=\{(x,y): x^2+y^2=1\}$ and that $\mathbb{R}\times\{0\}=\{(x,0): x\in\mathbb{R}\}$, then the situation is completely different and the right answer is $\mathbb{Z}*\mathbb{Z}$. In this case our space is path connected, so we can specify any base point we want (draw a picture, this is a circle with a line crossing it). we choose as a base point $x_1=(-1,0)$. So we may apply Seifert van Kampen here and the result follows after a mere group-theoretic computation. I guess you can see why $X$ and two circles with one common point are homotopy-equivalent, right?
