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So far, any source I consult will gladly talk about cobordism classes of closed (compact and without boundary) oriented manifolds, but I have yet to see an example of a pair of manifolds which are not cobordant.

As far as I can tell, every pair of one- or two-dimensional closed oriented manifolds are bordant, so such an example would seemingly only occur in dimensions 3 or higher. They reason I think this is that since the torus and the sphere are bordant, we can just "iterate" this cobordism in a suitable way and exhaust all possibilities by the classification theorem of compact oriented surfaces.

I was wondering if anyone knows of an example of a non-cobordant pair of closed oriented manifolds which is relatively easy to appreciate. If possible, an excplicit construction would be preferrable.

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up vote 4 down vote accepted

You can just consider some simple bordism variants.

If you're considering oriented bordism, then the first example appears when considering bordisms between $4$-manifolds. The signature $\sigma(X^{4k})$ of an oriented $4k$-manifold $X^{4k}$ is an oriented bordism invariant. Now $$\sigma(S^4) = 0$$ and $$\sigma(\Bbb C P^2) = 1,$$ so $S^4$ and $\Bbb C P^2$ are not oriented bordant. In fact, $\Omega_4^{\mathrm{SO}} \cong \Bbb Z$ with generator $[\Bbb C P^2]$.

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Why is the union of a cone of $S^4$ with a cone of $CP^2$ not a cobordism between them? – Catherine Ray Aug 3 '15 at 5:43
The cone on $\Bbb C P^2$ is not a topological manifold. In fact, the cone on any space $X$ that does not have the same integral homology as a sphere cannot possibly be a manifold, because for such spaces we have $\tilde{H}_\ast(X, X - \text{cone pt}; \Bbb Z) \not \cong \Bbb Z$. – Henry T. Horton Aug 8 '15 at 1:54

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