Compact manifold and eigenvalues Let $X$ be a compact manifold and $f : X \to X$ a smooth map.

Show that $\{x \in X | f(x) =x,\ 1 \text{ is not an eigenvalue of }df_x : T_xX \to T_xX\}$ is a finite set.

If $1$ isn't an eigenvalue then $\det(df_x - I)\neq 0$ which is an open set condition. So I guess we can do a "open cover has finite subcover" argument.
But I didn't really use $f(x)=x$ condition.
 A: This is false. Take the function $f: \Bbb R \to \Bbb R$ given by $f(x) = x + e^{-1/x^2} x^2 \sin(1/x)$. The function is chosen to have oscillating behavior near zero, while dampening it so that the resulting function is smooth. While this is a non-compact manifold, all that really matters here is the behavior near 0, so you can replace this with a function from the circle to itself which agrees with this function near a point on the circle.
There are fixed points at all $x = 1/k\pi$ for all integers $k$. Now take a derivative. For nonzero $x$ we get $$1 + \frac 2x e^{-1/x^2} \sin(1/x) + 2 e^{-1/x^2} x \sin(1/x) - \frac 1x e^{-1/x^2} \cos(1/x).$$
At $0$ the derivative (limit of the above expression) is 1.
For $x = 1/k\pi$ the derivative of the above expression evaluates to $1 - k\pi e^{-k^2 \pi^2} \cos(k\pi) \neq 1.$$
Therefore the set of fixed points so that $1$ is not an eigenvalue of $df_x$ is not closed: there are such fixed points which accumulate to 0, where 1 is an eigenvalue.
A: Caution This proves only that the set is discrete, without proving that the set is closed, this answer is incomplete/incorrect.
Since $X$ is compact, it suffices to show that the set $S$ is discrete.
If $S$ is not discrete, there is $s_0\in S$ and a sequence $s_n\in S$ so that $s_n\neq s_0$ for all $n$ and $s_n \to s$. Taking a subsequence if necessary, assume that $s_n$ lies in a coordinates chart $(U, \phi)$ which contains $s_0$. Then
$$ \phi \circ F\circ \phi^{-1}:\phi( U)\to \mathbb R^n$$
satisfies $F(x_0) = x_0$, $F(x_n) = x_n$ for all $n$, where $x_0 = \phi (s_0)$, $x_n = \phi(s_n)$. Taking a subsequence if necessary, assume
$$ v_n=\frac{x_n - x_0}{|x_n - x_0|}\to v\in \mathbb S^{n-1}$$
The Taylor expansion at $x_0$ gives
$$F(x) = F(x_0) + \nabla  F(x_0) \cdot (x-x_0) + o(|x-x_0|),$$
putting $x = x_n$, divide the above by $|x_n - x_0|$ and use $|x_n -x_0|^{-1} o(|x_n-x_0|)\to 0$, we obtain
\begin{align} v &= \lim_{n\to \infty} \frac{x_n - x_0}{|x_n -x_0|} \\
&= \lim_{n\to \infty} \frac{F(x_n)-F(x_0)}{|x_n - x_0|} \\
&= \lim_{n\to \infty} \nabla F(x_0) \frac{x_n - x_0}{|x_n - x_0|} \\
&= \nabla F(x_0) v.
\end{align}
Thus $\nabla F(x_0)$ has $1$ as an eigenvalue, which is a contradiction to $s_0\in S$. Thus $S$ is a discrete set.
