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Using the Sard-Smale theorem, it is relatively easy to show that Morse-Smale pairs on a manifold $M$ (i.e. pairs $(f,g)$ where $g$ is a metric on $M$, $f$ is a Morse function on $M$, and the stable/unstable disks corresponding to th flow of $-\nabla f$ intersect transversely) are dense in the appropriate topology. However, this argument requires the use of Banach manifolds. For the sake of a more accessible exposition, I was hoping to find an argument that uses only finite-dimensional methods, such as Thom transversality.

My thought was the following: Perhaps if we fix the metric $g$, we can show that functions $f$ such that $(f,g)$ is Morse-Smale are dense. (Just looking for a function seems to be more in the realm of classical applications of Thom transversality than also looking for a metric.) The Sard-Smale result certainly guarantees that this will be true for a generic $g$, but will it hold for any $g$? It would certainly seem very strange to me if there were a metric $g$ not admitting any such function $f$.

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The Sard-Smale result certainly guarantees that this will be true for a generic 𝑔, but will it hold for any 𝑔?

If $g$ is fixed, you can certainly use the Sard-Smale theorem to prove the existence of a Morse function $f$ so that the pair $(f,g)$ is Morse-Smale.

And yes, there's also a proof with less heavy machinery, see for instance Theorem 6.6 in the book "Lectures on Morse Homology" by Banyaga and Hurtubise.

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