# Fundamental group of a sphere with a handle

Consider a sphere $$S^2$$ with a circle attached to it by a point. I want to calculate its fundamental group. I will use the following different version of Van Kampen theorem

If $$X$$ is path connected and $$X=X_1 \cup X_2$$ with $$X_1$$ path connected and $$X_2$$ simply connected and both open and such that $$X_1 \cap X_2= X_0= A\cup B$$ where $$A,B$$ are disjoint and both open and simply connected. Then $$\pi_1(X)= \pi(X_1)* \mathbb{Z}$$

Then what I can do is cut a line from the handle to obtain $$X_1$$ and take a longer line as $$X_2.$$

In this way the intersection is given by two lines which are both simply connected.

Moreover the fundamental group of $$X_1$$ is trivial because if I cut a line from the handle then I can rectract the handle to a point on the sphere which is contractible.

Thus the $$\pi_1$$ of the sphere with a handle must be $$\mathbb{Z}.$$

Does it work?

You just want to show that any path in $$X$$ is homotopic to a path just in the circle $$S^1$$ (which we assume to be attached to $$S^2$$ at a point $$p$$). So take a path $$\gamma:[0,1]\to X$$. Then $$\gamma^{-1}(X\setminus S^1)$$ consists of finitely many open intervals. Consider such an interval $$(a,b)\subset[0,1]$$ that has $$\gamma(a)=\gamma(b)=p$$ and $$\gamma([a,b])\subset X$$. We know that $$S^2$$ is simply-connected, so $$\gamma|_{[a,b]}$$ is homotopic to the constant path at $$p$$. So $$\gamma$$ is homotopic to a path $$\tilde{\gamma}$$ with $$\tilde{\gamma}^{-1}(X\setminus S^1)$$ consisting of one less interval than $$\gamma^{-1}(X\setminus S^1)$$. By induction $$\gamma$$ is thus homotopic to a path entirely in $$S^1$$.
So we conclude $$\pi_1(X)\cong\pi_1(S^1)\cong\mathbb Z$$.