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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?

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There is no need to use Seifer-van-Kampen.

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 $\pi_1(X)\cong\pi_1(S^1)\cong\mathbb Z$.

In some sense we used exactly the idea of the proof of Seifert-van-Kampen.

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    $\begingroup$ I keep asking precise questions and people keep answering another question instead. I asked about my solution, not a different solution. $\endgroup$
    – ggeolier
    Oct 28 '20 at 21:19
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    $\begingroup$ I understand your frustration, but sometimes it is hard to answer "is my proof correct" question in any insightful way, especially when their proof is correct. @ggeolier $\endgroup$ Oct 28 '20 at 23:19

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