If $M$ is the connected sum of $n$ tori and $N$ is the connected sum of $n$ projective spaces, then what surface is $M \# N$?

Question

I don't know if this exercise is correct and I need your opinion:

Consider $M$ is the connected sum of $n$ tori and $N$ is the connected sum of $n$ projective spaces. What surface is $M \# N$? And if $N$ is the connected sum of $n$ Klein bottles?

My answer is the following:

In the first case, I think that it would be a connected sum of $3n$-projective spaces. And, in the second case, I think that it would be a connected sum of $4n$-projective spaces.

I have proved using that: $ T \# P^2(\mathbb{R}) \sim K \# P^2(\mathbb{R}) \sim P^2(\mathbb{R}) \# P^2(\mathbb{R}) \# P^2(\mathbb{R}) $ where $T$ is the torus, $K$ is the Klein bottle, and $P^2(\mathbb{R}) $ is the projective space.

I hope your answers.

Thanks.

Answer 1

Many such questions have been asked on stack exchange, so maybe I should just refer there to them. But I think what is sometimes not emphasised enough is that all such questions can be answered systematically using the following three facts.

1. The connected sum is non-orientable $\iff$ at least one of the factors in the connected sum is non-orientable.

2. The Euler characteristic is injective on the set of homeomorphism types of orientable compact surfaces. The Euler characteristic is injective on the set of homeomorphism types of non-orientable compact surfaces.

3. For any two surfaces $M,N$; $\chi(M \# N ) = \chi(M) + \chi(N) -2$.

So to identify a compact surface, you first check if any factors in the connected sum are non-orientable. Next you compute its Euler characteristic using the formula from 3. By 2 these two purely computational steps uniquely determine the surface.

What is more difficult is to prove fact 2, which is a consequence of the classification of surfaces.