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 $\pi_{1}(X) \approx *_{\alpha} \pi_{1}\left(A_{\alpha}\right) / N .$

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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..)

  • 1
    $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Nov 30, 2019 at 13:22
  • $\begingroup$ @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. $\endgroup$
    – Stephen
    Nov 30, 2019 at 13:31

1 Answer 1


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).

I hope this answers your question !

  • $\begingroup$ See the discussion at https:mathoverflow.net/questions/40945/, and also /175923, on the use of many base points. $\endgroup$ Jan 26, 2020 at 21:40

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