Nondegenerate fixed point

Let $M$ be a smooth $n$-manifold and let $f:M\to M$ be a smooth map with a fixed point $x$; identify $T_{f(x)}M$ with $T_{x}M$ along $f(x)=x$, so that the differential is an endomorphism $D{f}_x:T_{x}M\to T_{x}M$ (The differential of a smooth map). The fixed point $x$ is nondegenerate when $I-D{f}_x$ is an isomorphism of $T_{x}M$ (Invertible linear maps, linear isomorphisms, and inverse linear maps), equivalently when $1$ is not an eigenvalue of $D{f}_x$. A nondegenerate fixed point is isolated: in a chart at $x$ the displacement $u\mapsto u-\widehat{f}(u)$ has invertible derivative $I-D\widehat{f}(0)$ at $u=0$, hence is a local diffeomorphism near $0$ (The smooth inverse function theorem on manifolds) with the unique zero $0$ there, so a neighbourhood of $x$ contains no other fixed point. No choice principle is used, and $n=0$ is allowed (then $I-D{f}_x$ is an isomorphism of the zero space and the condition is vacuous).

• Convention. The condition is written $I-D{f}_x$, not $D{f}_x-I$; whether $1$ is an eigenvalue of $D{f}_x$ is insensitive to the order, since $\det (I-D{f}_x)=(-1{)}^n\det (D{f}_x-I)$ for an $n$-dimensional $T_{x}M$ and a determinant is nonzero exactly when the endomorphism is invertible. The sign matters for the value of the local index, not for nondegeneracy; see Isolated fixed point and local fixed point index and The index of a nondegenerate fixed point is the sign of det(I-Df).
• The two readings agree with graph transversality. The equivalence of the algebraic condition with transversality of the graph of $f$ to the diagonal of $M\times M$ at $(x,x)$ is Graph-diagonal transversality is exactly fixed-point nondegeneracy, and it is what brings nondegenerate fixed points under the intersection theory of the ambient manifold. Neither statement uses an orientation of $M$.