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In many places, Ricci flow with surgery is done with orientable manifolds. Morgan and Tian do not require orientability, but instead they impose the condition that $M^3$ have no embedded $\Bbb RP^2$ with trivial normal bundle. Of course, this holds when $M^3$ is orientable.

They need this to prove the strong canonical neighborhood assumption. For example, the manifold $\Bbb RP^2\times S^1$ with some asymmetric metric might develop a singularity along the $\Bbb RP^2$, not an $S^2$.

So, what if we do sugery along $S^2$s and $\Bbb RP^2$s? Does this work? It's easy enough to consider "nonorientable" $\varepsilon$-necks $\Bbb RP^2\times I$. A nonorientable standard solution is harder because $\Bbb RP^2$ is not the boundary of a compact 3-manifold (see here). Does this mean there's no hope?

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    $\begingroup$ You could add a cone over RP^2. I don't know how Ricci flow behaves with such singularities. $\endgroup$ Commented Oct 22, 2018 at 8:18

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