Getting acquainted with preimage orientation I'm trying to do this problem:
Let $f: S^2 \to \mathbb{R}^3$ given by $f(x,y,z)=z$. For the regular values $-1<t<1$, find the orientations of $f^{-1}(t).$
The hint is to find a positively oriented vector. The preimage orientation as follows:
If $f:X \to Y$ is transverse to $Z$, then, the orientation of $S =f^{-1}(Z)$ is given by the following relations:
$$N_x(S; X) \oplus T_x(S) = T_x(X)$$
and
$$df_x(N(S;X)) \oplus T_{f(x)}(Z) = T_{f(x)}(Y).$$
But I just don't know what to do with this. 
For instance, I can see that $T_x^(X)= (a,b,t)^\perp$ for $x=(a,b,t)$ with $a^2+b^2=1-t^2$, but I don't know what the orientation of this manifold is, or how to see it. 
Another general question I have is: how do we talk about whether a basis is positive or negative when talking about 2 dimensional subspaces of $\mathbb{R}^3$? We have two 3-dimensional vectors, and I don't see how we can say something about a linear map with positive determinant.
Any help with this and the above problem would be much appreciate.
 A: Second question first:
An orientation on a linear vector space can be given by a choice of ordered basis to be "positive". The standard orientation on $\mathbb{R}^n$ is given by the standard ordered basis $\{e_1,...,e_n\}$. If you are given a $2$-dimensional linear subspace $H \subset \mathbb{R}^3$, it does not automatically come with an orientation. To talk about "positive" or "negative" bases of it, you first need to choose such a basis. In some cases, an orientation on $H$ may be implicit in however it was given to you (e.g. as the preimage $f^{-1}(c)$ for some linear map $f: \mathbb{R}^3 \to \mathbb{R}$.)
Okay, first question. Fix $t \in (-1,1)$, let $C = f^{-1}(t) \subset S^2$ be the latitude line, and $p = (a,b,t) \in C$. Let's start by taking stock of all the orientations we know.
The image $\mathbb{R}$ carries the standard orientation, with $\{e_1\}$ a positive basis at every point.
$\mathbb{R}^3$ carries the standard orientation.
$S^2 \subset \mathbb{R}^3$ is probably assumed to carry the orientation given by an outward normal vector $n$ (this is standard). The equation
$$N_p(S^2) \oplus T_p(S^2) = T_p(\mathbb{R}^3)$$
tells us that an ordered basis $\{v_1,v_2\}$ for $T_p(S^2)$ is "positive" if the combined ordered basis $\{n, v_1, v_2\}$ is positive for $\mathbb{R}^3$. For example, one positive basis for $T_p(S^2)$ at $p = (a,b,t)$ (away from the poles $t = \pm 1$) is
$$v_1 = (-b, a, 0), \quad v_2 = \left(-at,-bt,a^2 + b^2\right).$$
Now we turn to the curve $C$. The second relation you give says
$$df_p(N(C; S^2)) \oplus T_t(\{t\}) = T_t(\mathbb{R}).$$
Since $T_t(\{t\})$ is trivial ($\{t\}$ is $0$-dimensional), this is just an isomorphism
$$df_p(N(C; S^2)) \cong T_t(\mathbb{R}).$$
Additionally, the restriction of $df_p$ to $N(C;S^2)$ is an isomorphism onto its image, so we actually have an isomorphism
$$df_p: N_p(C; S^2) \to T_t(\mathbb{R}).$$
A "positive" basis for $N_p(C;S^2)$ is one that corresponds, via this isomorphism, to a positive basis for $T_t(\mathbb{R})$. For example, we might take the previously defined
$$v_2 = \left(-at,-bt, a^2 + b^2\right)$$
as a positive basis vector, since $df_p(v) = (a^2 + b^2)e_1$ is a positive basis vector for $T_t(\mathbb{R})$.
Now return to the first relation.
$$N_p(C;S^2) \oplus T_p(C) = T_p(S^2)$$
tells us that a vector $w \in T_p(C)$ forms a positive basis iff $\{v_2, w\}$ forms a positive basis for $T_p(S^2)$. For example, we might take $w = -v_1$, since we know that $\{v_1, v_2\}$ is a positive basis.
That's it! $C = f^{-1}(t)$ is oriented in the $-v_1 \in T_p(C)$ direction at each point. In plain English, $C$ is oriented going clockwise around the positive $z$-axis. Or, thinking of $S^2$ as a globe and $C$ as a latitude line, the preimage orientation runs east to west.
