# Van Kampen's Theorem: clarification about the basepoint $x_0$

van Kampen's Theorem: (as formulated in Allen Hatcher's book, p.43)

If $$X$$ is the union of path-connected open sets $$A_{\alpha}$$ each containing the basepoint $$x_{0} \in X$$ and if each intersection $$A_{\alpha} \cap A_{\beta}$$ is path-connected, then the homomorphism $$\Phi: *_{\alpha} \pi_{1}\left(A_{\alpha}\right) \rightarrow \pi_{1}(X)$$ is surjective. If in addition each intersection $$A_{\alpha} \cap A_{\beta} \cap A_{\gamma}$$ is path-connected, then the kernel of $$\Phi$$ is the normal subgroup $$N$$ generated by all elements of the form $$i_{\alpha \beta}(\omega) i_{\beta \alpha}(\omega)^{-1}$$ for $$\omega \in \pi_{1}\left(A_{\alpha} \cap A_{\beta}\right),$$ and hence $$\Phi$$ induces an isomorphism $$\pi_{1}(X) \approx *_{\alpha} \pi_{1}\left(A_{\alpha}\right) / N .$$



Question:

In the application of the theorem, can we freely choose any point in $$X$$ as the basepoint? Or should we consider each possible basepoint, separately?

For example, if I have a cover of $$X$$ consisting of the subspaces $$U\cup A$$ and $$(U \cup C) \setminus B$$, for some non-empty subspaces $$A$$, $$B$$ and $$C$$, then any basepoint $$x_0 \in B$$ does not belong to the subspace $$(U \cup C) \setminus B$$.
Does this mean that my choice of cover only allows me to find the fundamental group $$\pi_1(X, x_0)$$ w.r.t. to each basepoint $$x_0\in (U\cup A) \cap ((U \cup C) \setminus B)$$, but not the fundamental group $$\pi_1(X, y_0)$$ w.r.t. to a basepoint $$y_0 \notin (U\cup A) \cap ((U \cup C) \setminus B)$$?

And so, to find $$\pi_1(X, y_0)$$ I should consider a different cover of $$X$$ that has $$y_0$$ in the intersection of the subspaces it contains?

(Hopefully my question makes sense..)

• van Kampen's theorem applies to any basepoint $x_0 \in \bigcap A_\alpha$. If $x_0 \notin A_\alpha$ for some $\alpha$, then $\pi_1(A_\alpha,x_0)$ is not defined and thus the statement doesn't make sense. Nov 30, 2019 at 13:22
• @PaulFrost Yes, of course. My question was about whether or not the fundamental group we find, based on a given basepoint, remains the same even when we change the basepoint. Popyaitte has answer the question, though. Thanks again. Nov 30, 2019 at 13:31

Your example is not totally clear but I think I see what you missed. In Hatchers statement you notice he only writes $$\pi_1(X)$$ and not $$\pi_1(X,x_0)$$. This due to the fact that his hypothesises imply that $$X$$ is path connected.
This last thing implies that the fundamental group at each point is the same (up to isomorphism) : Suppose that $$X$$ is path connected one has a clear isomorphism between $$\pi_1(X,x_0)$$ and $$\pi_1(X,y_0)$$ for all $$x_0,y_0 \in X$$. Just choose a path $$\gamma : [0;1]\longrightarrow X$$ such that $$\gamma(0)=x_0$$ and $$\gamma(1)=y_0$$. Then you have the isomorphism $$\phi_\gamma : \pi_1(X,x_0)\longrightarrow \pi_1(X,y_0)$$ given by $$\phi_\gamma([l]) = [\gamma * l * \gamma^{-1}]$$.
So finally, even if you have to take care that such a point $$x_0$$ and a covering of $$X$$ path-connected sets of the form $$(A_\alpha)_\alpha$$ as in the theorem exist in order to apply it, since each set involved in the statement is path connected and contains $$x_0$$ you can talk about the fundamental group of such a set (which is the fundamental group of this set at $$x_0$$ a priori but doesn't depend on the choice of the point by path-connectedness, so it depends only on the set considered).