Compute $\pi_1(S^1 \vee S^1)$
My attempt :$\pi_1(S^1 \vee S^1,x)\cong\mathbb{Z}*\mathbb{Z}$
My question: How to compute $\pi_1(B)$(see the given figure) ?
My attempt : This can be shown by the diagram 
from the given diagram B retract to the circle
But I want mathematical proof
We can show that $B$ and $S^1$ are homotopy equivalent, i.e. There is a map $f:B\to S^1$ and a map $g:S^1\to B$ such that $f\circ g\simeq Id_{S^1}$ and $g\circ f\simeq Id_B$.
$B$ can be expressed as $([-1,1],0)\lor (S^1,s_0)$. Consider $f:B\to S^1$ defined by $f(t,s_0)=s_0$ for all $t\in [-1,1]$ and $f(0,s)=s$ for all $s\in S^1$. And also $g:S^1\to B$ defined by $g(s)=(0,s)$for all $s\in S^1$.
Observe that $$f(g(s))=f(0,s)=s,\ \forall s\in S^1$$ ($f\circ g = Id_{S^1}$ and therefore $f\circ g \simeq Id_{S^1}$ ), $$g(f(t,s_0))=g(s_0)=(0,s_0)$$ and $$g(f(0,s))=g(s)=(0,s).$$ So we only need to prove that $g\circ f\simeq Id_B$. To do so consider the following homotopy: $$H:B\times I\to B,\quad H((t,s_0),\lambda)=(\lambda t,s_0),\quad H((0,s),\lambda)=(0,s).$$
We can see that $H((t,s_0),0)=(0,s_0)$ and $H((0,s),0)=(0,s)$ so we have $H_0=g\circ f$. Also, $H((t,s_0),1)=(t,s_0)$ and $H((0,s),1)=(0,s)$, thus $H_1=Id_B$. Therefore we can conclude that $g\circ f\simeq Id_B$.
As homotopy equivalences induces a isomorphism between homotopy groups, we have $$\pi_1(B)\simeq \pi_1(S^1)=\mathbb{Z}.$$