Question: Let f:X -> X be a map with fixed points x, i.e. f(x) = x. If 1 is not an eigenvalue of df_x: T_x(X) ->

Let f:X -> X be a map with fixed points x, i.e. f(x) = x. If 1 is not an eigenvalue of df_x: T_x(X) -> T_x(X), then x is called a Lefschetz fixed point of f. f is called a Lefschetz map if all its fixed points are Lefschetz. Prove that if X is compact and f is Lefschetz, then f has only finitely many fixed points

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