Calculate the fundamental group of $S^1 \cup (\Bbb{R}_{\geq 0}\times\{0\})$
Ok, so Im starting to get the intuition in how to solve this kind of problems, but whenever I try to do it somehow more formally, I get stuck.
So I started by defromation retracting $S^1 \cup (\Bbb{R}_{\geq 0}\times\{0\})$ to $S^1 \cup [0,1]$. And my intuition says that the latter is homotopically equivalent to $S^1 \vee S^1$. By somehow identifying $S^1$ with $I/0\sim1$. The problem is im nowhere close to writing this.
I thought of something like: $f:S^1 \cup [0,1] \to S^1 \vee S^1$
$$f(x) = \begin{cases} x & \quad \text{if } x \in S^1 \\ e^{2\pi i t} & \quad \text{if } t \in [0,1] \end{cases}$$
Which is continuous by the pasting lemma, and that it's constant in the classes of the relation $0 \sim 1$. However, I don't know how to say that they go to the different $S^1$ of the wedge. Is this the way? Is there another way?
A good picture is worth a thousand words. Presuming that $S^1$ is the usual unit circle in $\mathbb{R}^2$, then the set you are describing is pictured below:
The horizontal ray (i.e. the set $\mathbb{R}_{\ge 0} \times \{0\}$) retracts to a point (in particular, to the point $(1,0)$), and you are left with the circle.