Show that the projection map of $S^n_+$ is orientation preserving $\iff$ $n$ is even This is problem $22.10(b)$ from Introduction to Manifolds (Tu). This question has been posed here and here, but no proof has been given.
The text provided the answer to $22.10(a)$, which is $(-1)^n dx^1 \wedge \ldots \wedge dx^n$.

We must first show that $\pi$ is a diffeomorphism. I get stuck here. I want to show that
$\sigma : U \rightarrow \mathbb{R}^n \times \{0\}$, defined by
$$\sigma(x^1, \ldots, x^{n+1}) = (x^1, \ldots, x^n, 0), $$
is a diffeomorphism. But the Jacobian is clearly not invertible.
If I could show that $\sigma$ was a diffeomorphism, then I would finish the proof as follows. Since $\mathbb{R}^n\times \{0\} \simeq \mathbb{R}^n$, we now only consider the latter. A nowhere-vanishing form on $\pi(U) \subset \mathbb{R}^n$ is $dx^1 \wedge \ldots \wedge dx^n$, and its pullback by $\pi$ is the same expression. Therefore, $\pi$ is orientation preserving $\iff$ $n$ is even. $\square$
 A: Even after you prove that $\pi$ is a diffeomorphism, there is a big hole in your proof: why is the pullback by $\pi$ of ${\rm d}x^1 \wedge \cdots \wedge \cdots \wedge {\rm d}x^n$ the same expression? Actually,
$$\pi^{*} ({\rm d}x^1 \wedge \cdots \wedge \cdots \wedge {\rm d}x^n) = \det({\rm d} \pi) {\rm d}x^1 \wedge \cdots \wedge \cdots \wedge {\rm d}x^n $$
And you have not computed ${\rm d} \pi$ (actually I suspect you have but you probably went about it the wrong way). Remember that we're considering $U$ as an intrinsic manifold, and so to compute the matrix of ${\rm d} \pi$ we must use charts. But first, why is $\pi$ a diffeomorphism? Well, clearly the map
$$\mathbb{R}^2 \ni (a, b) \mapsto \left(a, b, \sqrt{1 - a^2 - b^2} \right) \in U$$
is smooth and it is the inverse of $\pi$. Therefore $\pi$ is a diffeomorphism. Now let's calculate the matrix of ${\rm d} \pi$. An appropriate chart to use for $U$ would be the stereographic projection $\varphi$ from the south pole (which covers $U$), given by:
$$x = (x^1, \cdots, x^{n+1}) \in U \subset \mathbb{S}^n \mapsto \varphi(x) =  \frac{1}{1 + x^{n+1}}(x^1, \cdots, x^n) \in \mathbb{R}^n$$
A straightforward calculation shows that the inverse of this map is given by:
$$\mathbb{R}^n \ni u = (u^1, \cdots, u^n) \mapsto \psi(u) =  \left(\frac{2u^1}{ \| u \|^2 + 1}, \cdots,  \frac{2u^n}{ \| u \|^2 
+ 1}, \frac{ 1 - \|u \|^2}{ \|u \|^2 + 1}\right) \in \mathbb{S}^n$$
Consider the coordinate representation of $\pi$, given by $\hat{\pi} = \pi \circ \varphi^{-1} : \varphi(U) \subset \mathbb{R}^2 \to \mathbb{R}^2$. Almost by definition, the matrix of ${\rm d} \pi$ is just the matrix of ${\rm d} \hat{\pi}$. Now, since the map $\hat{\pi}$ is just
$$x = (x^1, \cdots, x^n) \mapsto \left(\frac{2x^1}{\|x\|^2 + 1}, \cdots,  \frac{2x^n}{\|x\|^2 + 1}\right)$$
we see that the $(i, j)$-th entry of the matrix is given by:
$$\frac{\partial}{\partial x^j}\left(\frac{2x^i}{\|x\|^2 + 1} \right) = \frac{ 2 \delta^{i}_{j}(\| x\|^2 + 1) - 4 x^i x^j}{(\|x\|^2 + 1)^2}$$
At first glance you might ask: alright, but how on earth do we calculate the determinant of this matrix? Alas, all is not lost: we don't need to! All that we're interested in is the sign of $\det({\rm d} \hat{\pi})$. And since $\hat{\pi}$ is a diffeomorphism and our domain is connected, the sign of $\det({\rm d} \hat{\pi})$ is constant. In particular, we may choose whatever convenient point we want to in order to make the calculations easier. So we can just choose the north pole $N$. And there, we see that:
$${\rm d } \pi_N = 2 \operatorname{Id}$$
where $\operatorname{Id}$ is the identity matrix of dimension $n$. Therefore, the sign of $\det({\rm d} \pi)$ is always positive. So, the orientation forms
$$\pi^{*} ({\rm d}x^1 \wedge \cdots \wedge \cdots \wedge {\rm d}x^n)  = 2^n {\rm d}x^1 \wedge \cdots \wedge \cdots \wedge {\rm d}x^n$$
and
$$(-1)^n {\rm d}x^1 \wedge \cdots \wedge \cdots \wedge {\rm d}x^n$$
only ever induce the same orientation when $n$ is even, as desired.

Let's verify that $\varphi$ and $\psi$ are indeed inverses. First we need to verify that $\psi$ is well defined. Indeed, since
$$
\|\psi(u) \|^2 = \left(\sum_{1 \leq i \leq n} \frac{4 u_i^2}{(\|u\|^2 + 1)^2} \right) + \frac{(1 - \|u \|^2)^2}{(\|u \|^2 + 1)^2} = \frac{4 \|u \|^2 + \|u\|^4 - 2 \|u \|^2 + 1}{\|u\|^4 + 2 \|u \|^2 + 1} = 1 \ \forall u \in \mathbb{R}^n
$$
then $\psi$ is well defined. Now, if $u = (u^1, \cdots, u^n) \in \mathbb{R}^n$, then the point $$(\varphi \circ \psi)(u^1, \cdots, u^n) = \varphi\left(\frac{2 u^1}{\|u \|^2 + 1}, \cdots, \frac{2 u^n}{\|u\|^2 + 1}, \frac{1 - \|u \|^2}{\|u \|^2 + 1} \right)$$ is just
$$\begin{aligned}&\frac{1}{1 + \frac{1 - \|u \|^2}{\|u\|^2 + 1}} \cdot \frac{ 2}{\|u\|^2 + 1 }(u^1, \cdots, u^n)\\ &= \frac{\|u\|^2 + 1}{2} \cdot \frac{ 2}{\|u\|^2 + 1 } (u^1, \cdots, u^n) 
 \\ &= u \end{aligned}$$
so $\varphi \circ \psi = \operatorname{Id}_{\mathbb{R}^n}$. We must now also show that $\psi \circ \varphi = \operatorname{Id}_{\mathbb{S}^n}$. So let $v = (v^1, \cdots, v^{n+1}) \in \mathbb{S}^n$. To calculate $$(\psi \circ \varphi)(v) = \psi\left(\frac{1}{1 + v^{n+1}}(v^1, \cdots, v^n) \right)$$ we must first calculate $\| \hat{v}\|^2 + 1$, where $\hat{v} = \varphi(v)$. Now, since $v \in \mathbb{S}^n$, we have that:
$$\| \hat{v} \|^2 = \frac{(v^ 1)^2 + \cdots + (v^n)^2}{(1 + v^{n+1})^2} = \frac{1 - (v^{n+1})^2}{(1 + v^{n+1})^2}$$
and therefore
$$\| \hat{v} \|^2  + 1 = \frac{2}{ 1+ v^{n+1}}$$
so that if $1 \leq i \leq n$, the $i$-th coordinate of $(\psi \circ \varphi)(v)$ is just $$2 \cdot \frac{v^i}{1+v^{n+1}} \cdot \frac{1 + v^{n+1}}{2} = v^i$$
And a similar calculation shows that the $(n+1)$-th coordinate of $(\psi \circ \varphi)(v)$ is just $v^{n+1}$ too. So indeed $\psi \circ \varphi = \operatorname{Id}_{\mathbb{S}^n}$.
