Isolatedness of Lefschetz map - almost there $f$ 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.
 A: Suppose $x_0$ is a Lefschetz fixed point. Take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood (after composing with proper coordinate functions) I can 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 $f(0)=0.$ Consider the function $g=f-id.$ Then $g(0)=0$ and by lefschetz condition $det(dg)(0)\neq 0$ (as 1 is not an eigenvalue of $df$). Hence $g$ is a local diffeomorphism by inverse function theorem and we are done.
Just out of curiosity are you reading the book differential topology by Guillemin and Pollack.  
A: An alternative solution, directly continuing Jellyfish's beginning of a solution to G&P's Problem 5-10, Chapter 1.
Since $df_x$ has no eigenvalues equal $1$, by Problem 5-9 of G&P
$$\text{graph}(df_x) \pitchfork \Delta(T_x(X)\times T_x(X)).$$
It's an easy exercise to check that $\Delta(T_x(X)\times T_x(X)) = T_{x,x}(\Delta(X\times X)),$ so we now have 
$$\text{graph}(f) \pitchfork \Delta(X\times X).$$
 The overlap of these two sets is $W := \{(x,f(x))\colon x\in X, x=f(x)\},$ in bijective (in fact, diffeomorphic) correspondence with the set of fixed points of $f$. Now, by the transversality theorem from G&P, $W$ is a submanifold of $X\times X$. Furthermore, $\text{codim}(W) = \text{codim}(\text{graph}(f)) + \text{codim}(\Delta(X\times X)) = 2n.$ But this shows that $W$ is in fact a manifold of zero dimension, that is a set of isolated points. Thus the set of fixed points of $f$ is also an isolated set. As a subset of compact $X$ it must thus be finite.
