Orientation reversing diffeomorphism for Manifolds

Question

So i just read the about the connected sum of two connected manifolds.

Given two connected $m$-dimensional manifolds $M_1,M_2$, let $h_i: \mathbb{R}^m \to M_i, i = 1,2$ be two embeddings. If both manifolds are oriented, then we assume that $h_1$ preserves the orientation and $h_2$ reverses it.

I'm quoting from Kosiniski's Differential Manifolds:

Let $\alpha:(0,\infty) \to (0, \infty)$ be an arbitrary orientation reversing diffeomorphism. We define $\alpha_m: \mathbb{R}^m - \boldsymbol{0} \to \mathbb{R}^m - \boldsymbol{0}$ by $$ \alpha_m(v) = \alpha(\vert v \vert)\frac{v}{\vert v \vert}.$$

The connected sum $M_1 \# M_2$ is the space obtained from the disjoint union of $M_1 - h_1(\boldsymbol{0})$ and $M_2-h_2(\boldsymbol{0})$ by identifying $h_1(v)$ with $h_2(\alpha_m(v))$

I don't understand the orientation reversing diffeomorphism $\alpha$. My questions are:

1. How do i need to imagine $\alpha$ to operate? There is no other information but i was wondering, how or why $\alpha$ is an orientation reversing map in the first place other than being told it is. Why does an orientation reversing diffeomorphism (in this particular case) always exist?

I know that if $(v_1,v_2,...,v_m)$ is an ordered basis of $\mathbb{R}^m$, then $(-v_1,v_2,...,v_m)$ is an ordered basis with the reversed orientation.

Is that what $\alpha$ does? If so, why is $\alpha$ mapping from $(0,\infty)$ to $(0,\infty)$?

2. In addition to my first question by looking at $\alpha_m$, how is $\alpha_m$ operating on the given Manifolds?

3. Given an oriented connected smooth $n$-manifold $M$. Is there any canonical way to find an embedding $h: \mathbb{R}^n \to M$ that reverses the orientation of $M$?

Answer 1

1. If you want an example of $\alpha$, just choose $\alpha(x):=\frac{1}{x}$. Its derivative is always negative, meaning it is orientation-reversing. Many other strictly decreasing functions $\alpha:(0,\infty)\to(0,\infty)$ works. They just have to satisfy $$\lim\limits_{x\to0^+}\alpha(x)=+\infty,\ \lim\limits_{x\to+\infty}\alpha(x)=0$$ and be smooth and have a derivative that is negative everywhere (so they would be strictly decreasing). You may see that $\alpha(x)=\frac{1}{x^n}$ also satisfy the condition, for each positive integer $n$.

Derivative of a smooth map $F$ at a point $x$ can be seen as a linear map from the tangent space $T_x\Bbb R$ to the tangent space $T_{f(x)}\Bbb R$. You may view both $T_x\Bbb R$ and $T_{f(x)}\Bbb R$ as the same as the vector space $\Bbb R$ itself, and the linear map would be defined by $v\mapsto F'(x)v$. Whether or not this linear map is orientation-preserving or -reversing just depends on whether $F'(x)$ is positive or negative in this one-dimensional case.

2. $\mathbb{R}^m - \boldsymbol{0}$ can be seen as the union of many lines radiating from $\boldsymbol{0}$, each line is of the form $\{cv:v\text{ is a fixed unit vector}, c\in(0,\infty)\}$. You may see that each of such line looks exactly like $(0,\infty)$, just pointing in some directions other than along x-axis. The function $\alpha_m$ flips each of these lines the same way $\alpha$ flips $(0,\infty)$. The far points are mapped close to $\boldsymbol{0}$, points close to $\boldsymbol0$ are mapped to far away, and all points on a line $\{cv:v\text{ is a fixed unit vector}, c\in(0,\infty)\}$ remain on the same line after the map $\alpha_m$.

3. Sorry but I don't understand your question. If you have two oriented manifolds $M_1,M_2$ of the same dimension, you may try to find a smooth map $F:M_1\to M_2$ whose derivative's determinant is negative. But if you are talking about reversing the orientation of just one oriented manifold $M$, you should be looking for a diffeomorphism $F:M\to M$, not function from some other manifold like $\Bbb R^n$.