I must solve this exercise:
$S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces which are connected and compact. Is it correct that $S_1\#S_2$ is not homeomorphic to $S_3\#S_4$ ?
I tried to find a counterexample in order to prove that it is not correct. My approach is the following:
- I choose a surface $S_1$
- Classification theorem says that $S_1$ is homeomorphic to a sphere or to a connected sum of tori or to a connected sum of projective planes.
- I decide $S_1$ to be homeomorphic to a sum of three projective planes: $S_1 \cong \mathbb{P} \# \mathbb{P} \#\mathbb{P}$
- I take $S_2$ = sphere, $S_3$ = torus T, $S_4$ = projective plane $\mathbb{P}$
- I have four pairwise non homeomorphic functions, but $S_1 \# S_2 =S_1$ and $T\#\mathbb{P} \cong \mathbb{P} \#\mathbb{P} \#\mathbb{P}$ so $S_1 \# S_2 \cong T\#\mathbb{P}$
In this way I found a counterexamle which proved that there exists a case in which $S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces but $S_1 \# S_2 \cong T\#\mathbb{P}$ ($A \implies \neg B$). So the question in the exercise is false.
Is my answer right?