Is it true that if $M$ is a compact manifold and $X,Y$ are submanifolds of $M$ which intersect transversely that the intersection $X\cap Y$ consists of finitely many points?
I'm trying to understand a proof of the Lefschetz fixed point theorem which, so far as I can tell, makes implicit use of this fact.
My immediate impression is that the answer is yes because otherwise what good is compactness here? I'm trying to prove it by arguing that if there were infinitely many points of intersection then there would be an open cover with no finite subcover and I keep getting stuck. Is there a better way to proceed or is the result simply wrong?
I don't necessarily need a full proof; a yes or no and if yes a push in the right direction would be more than appreciated already.
Here's an example that satisfies your hypotheses: the 2-manifold $M = S^2 =$ the one point compactification of $\mathbb{R}^2$, and the 1-dimensional submanifolds $X,Y \subset \mathbb{R}^2 \subset S^2$ given by $X = \{(t,0) \,\, | \,\, t \in (0,1)\}$, and $Y = \{(t,\sin(1/t) \,\, | \,\, t \in (0,1)\}$.