$f$ is a Lefschetz map on a compact manifold X. And I need to show the Lefschetz fixed point is isolated.
I proved that the graph of f is transversal to the diagonal inside $X \times X$, then I don't know how to proceed from here. Thank you very much for your help.
Suppose $x_0$ is a Lefschetz fixed point. Then we know that given 1 is not an eigenvalue of $df$ at $x_0$, take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood, think of $f$ as a map from open ball in $\mathbb{R}^n$ (say $B$), to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$ such that $f(0)=0.$ Then $det(df - id)(0)\neq 0$. Hence the preimage of $f-id$ is a local diffeomorphism.
But how can I show that it is 0-manifold? (Then 0-manifold has discrete topology, so $x_0$ is isolated.) I feel there is some trivial fact leads me to the conclusion, but I couldn't get it. Would someone be able to help point out? Thank you!
If $g$ is a diffeomorphism then $g$ is bijective in that small neighborhood. If $f$ has another fixed point a in that neighborhood then $g(a)=0$. This will contradict the bijcetivity of $g$. Therefore the Lefschetz fixed point is a 0-manifold. In other word, isolated.