# Regarding orientation and orientation-reversing in local diffeomorphism

I am confused about orientation and orientation reversing in local diffeomorphism $f$ from manifold $X$ to $Y$ at some points. So, what does $f$ orientation-reversing at a point mean?

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## 2 Answers

$X$ and $Y$ are oriented, which means every tangent space $T_xX$ and $T_yY$ has an orientation. $f$ is orientation-preserving (-reversing) at $x$ if and only if the linear map $df_x\colon T_xX\to T_{f(x)}Y$ preserves (reverses) orientation.

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To elaborate on Ted's answer, to say that $X$ and $Y$ are oriented manifolds is to impose on the tangent spaces $T_x(X)$ and $T_{f(x)}(Y)$ a fixed, smooth choice of orientation at any given $x \in X$ (recall that an orientation of, say, $T_x(X)$ is just a choice of ordered basis $\alpha_1, \ldots, \alpha_n$, and likewise for $T_{f(x)}(X)$).

Now, $f$ is a local diffeomorphism at $x$, so the differential $df_x: T_x(X) \rightarrow T_{f(x)}(Y)$ is an isomorphism of vector spaces, and hence must take a positively-oriented basis $\alpha_1, \ldots, \alpha_n$ for $T_x(X)$ to a basis $B' = \beta_1', \ldots, \beta_n'$ for $T_{f(x)}(Y)$.

Since $Y$ is oriented, we already have a positively oriented $B = \beta_1, \ldots, \beta_n$ on $T_{f(x)}(Y)$. The matrix corresponding to the linear automorphism of $T_{f(x)}(Y)$ taking the ordered basis $B$ to $B'$ then has either positive or negative determinant. If the determinant of this matrix is positive, $B'$ is also a positively-oriented basis for $T_{f(x)}(Y)$, and we say that $f$ is orientation-preserving; if the determinant is negative, $B'$ is a negatively-oriented basis, and we say that $f'$ is orientation-reversing.

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