Set consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into additive group?

Question

How do I see that the set $\mathfrak{N}_n$ consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into an additive group?

Answer 1

A natural operation is disjoint union. If $[M]$, $[N]$ are cobordism classes of $M, N$ respectively then define the sum to be $[M] + [N] = [M \sqcup N]$, the cobordism class of the disjoint union of the two.

This is clearly associative, commutative and has identity as the null manifold class $[\emptyset]$. $[M] + [M] = [M \sqcup M]$, and as $M \sqcup M$ bounds $M \times [0, 1]$ (in the unoriented category!), $2[M] = 0$, so a cobordism class is it's own inverse.

The resulting additive group is denoted as $\Omega_n^{un}$, called the $n$-th unoriented cobordism group of a point. A classic question in topology is to compute these groups. Some examples are $\Omega_0^{un} \cong \Bbb Z/2$, $\Omega_1^{un} \cong 0$, $\Omega_2^{un} \cong \Bbb Z/2$, $\Omega_3^{un} \cong 0$, $\Omega_4^{un} \cong \Bbb Z/2 \times \Bbb Z/2$ and $\Omega_5^{un} \cong \Bbb Z/2$.