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Let $M$ be an $n$-dimensional manifold and let $[M]$ denote the unoriented bordism class of $M$. Forming the usual commutative graded ring $\text{MO}_n$ we know that $$\text{MO}_* \simeq \mathbb{Z}_2[u_n : n \neq 2^r -1, \deg u_n = n]$$ as graded rings.

Writing out the first groups we have that $\text{MO}_0 = \mathbb{Z}_2$, $\text{MO}_1 = 0$, $\text{MO}_2 = \mathbb{Z}_2$ (generated by $[\mathbb{R}P^2]$), $\text{MO}_3 = 0$ and so on.

From this I can see that $\mathbb{R}P^2$ is not the boundary of some $3$-dimensional compact manifold. But is it true that any closed 3-dimensional manifold is the boundary of some 4-dimensional compact manifold? That is, does being cobordant to a manifold which bounds imply that the manifold itself bounds?

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up vote 2 down vote accepted

Cobordism is an equivalence relation, if $L$ is cobordant to $M$ and $M$ is cobordant to $N$, $L$ is cobordant to $N$ (just glue to cobordisms together). In particular, $M$ is zero-cobordant iff it's a boundary (again, if $M$ is cobordant to $N$ and $N=\partial W$, just glue the cobordism and $W$ together to get a manifold which is bounded by $M$).

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The answer is yes.

Your question is in fact about the transitivity of the cobordism relation: The empty set is considered as an n-manifold for every n. You have a manifold M cobordant to a manifold N which bounds, i.e. is cobordant to $\emptyset$. So from transitivity it follows that M is also cobordant to $\emptyset$.

The proof of the transitivity of the cobordism relation is not hard (but you will need the collar neighborhood theorem in the smooth category), try it on your own!

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