2

i had the exercise to compute the fundamental group of the torus minus one point p.

I know that the fundamental group of the torus is $\pi_1(T^2) = \pi_1(S^1) \times \pi_1(S^1) = \Bbb Z \times \Bbb Z$. So:

$U :=$ open neighborhood of p

$V := T^2 \backslash \{p\}$

Then U and V are both path-connected and open in $T^2$, U $\cap$ V is path connected and $T^2 = U \cup V$.

So from my understanding we should have: $\pi_1(T^2) = \pi_1(U) * \pi_1(V)$ (where i use $*$ for free group)

but since $\pi_1(U) = 0$ (because U is contractible) this implies $\Bbb Z \times \Bbb Z = \pi_1(T^2) = \pi_1(V) = \pi_1(T^2 \backslash \{p\})$

But i looked it up and $\pi_1(T^2 \backslash \{p\}) = \Bbb Z * \Bbb Z $ which is not the same as $\Bbb Z \times \Bbb Z$, right? Does somebody know where i made a mistake?

Ton910
  • 145

1 Answers1

3

The mistake is that kn Van Kampen you don't get the free product, you get the amalgamated product $\pi_1(T^2\setminus\{p\})*_{\pi_1(U\cap V)} \pi_1(U)$

Here, $U\cap V$ retracts onto a circle so it has nontrivial $\pi_1$, so in the amalgamated product it will kill $[a,b]$ in $\mathbb Z* \mathbb Z$, that's why the torus has $\pi_1$ $\mathbb Z^2$.

To get the correct result, I think you have to get a geometric handle on $T^2\setminus \{p\}$ (find a nice space which it is homotopy equivalent to)

Maxime Ramzi
  • 43,598
  • 3
  • 29
  • 104