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Let X be a Hausdorff space such that X is the union of A and B; where both A and B are homeomorphic to a torus. The intersection of A and B is a singleton. I want to compute the fundamental group of X.

Intuitively/non-rigorously, I can say the fundamental group of X is the free product of the fundamental group of A and that of B. I don't see why the hausdorffness condition is necessary. Can anyone please tell me why we need the hausdorffness condition to apply Van Kempen theorem to solve this problem? Thank you so much.

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The key idea is that since $X$ is Hausdorff, any compact subspace of it is closed, and so $A$ and $B$ are both closed in $X$. This gives you a handle on the topology of $X$ and not just the topologies of $A$ and $B$. In particular, you can cover $X$ with open sets which deformation-retract to $A$ and $B$ (and whose intersection is contractible), which is what you need in order to use van Kampen. For instance, you can take $U$ to be $A$ together with a small open disk around the intersection point in $B$ and $V$ to be $B$ together with a small open disk around the intersection point in $A$. (Exercise: use the fact that $A$ and $B$ are both closed in $X$ to prove that $U$ and $V$ really are open in $X$.) To prove that $U$ really deformation-retracts to $A$, for instance, you can again use the fact that $A$ and $B$ are closed in $X$, so that by the pasting lemma you can check that a homotopy is continuous by just checking that it is continuous separately on $A$ and $B$.

Note that if $X$ is not Hausdorff, then all sorts of horrible things can happen. For instance, letting $T$ be a torus and $S$ be $\{0,1\}$ with the indiscrete topology, then $X$ could be $T\times S\setminus\{(x,1)\}$ for some point $x\in T$. You can then write $X$ as the union of subspaces $A=T\times\{0\}$ and $B=(T\setminus\{x\})\times\{1\}\cup\{(x,0)\}$ which are each homeomorphic to $T$ and intersect only at $(x,0)$. In this case the projection $X\to T$ is actually a homotopy equivalence, so $\pi_1(X)$ is not the free product of $\pi_1(A)$ and $\pi_1(B)$. And this is one of the tamer examples of what could happen if $X$ is not Hausdorff; there are many other more complicated possibilities.

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    $\begingroup$ Thanks so much, Eric Wofsey. It's really really helpful. I wonder how someone can come up with such an elegant counter-example! Thank you again. $\endgroup$ – MMM Oct 2 at 6:24
  • $\begingroup$ Another thing: I did not know how to accept an answer. Now I know it. Thank you again. Please accept my apology. $\endgroup$ – MMM Oct 6 at 6:46

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