# Oriented manifolds and the intermediate value theorem

Working through a proof that any connected orientable manifold has two orientations (From Tu's manifolds), I'm having a difficulty with a small nuance in the proof. Consider a manifold $$M$$ with two orientations $$\mu$$ and $$\nu$$, and some arbitrary point $$p \in M$$. $$\mu_p$$ and $$\nu_p$$ are the same or opposite orientations of $$T_p M$$. The strategy of the proof is to show that the function $$f:M \rightarrow \{1,-1\}$$ (where $$f(p) = 1$$ if $$\mu_p = \nu_p$$ and $$f(p) = -1$$ if $$\mu_p = - \nu_p$$) is locally constant, and therefore globally constant since $$M$$ is connected.

Okay, all well and good. Now, the orientations are continuous, so for any point there is a connected open neighborhood of $$p$$, $$U$$, where $$\mu = [(X_1,...,X_n)]$$ and $$\nu = [(Y_1,...,Y_n)]$$ are equivalence classes of continuous frames on $$U$$. Of course, there is a transition matrix between these frames, $$A:U \rightarrow GL(n,\mathbb{R})$$ with continuous entries, meaning the nonsingular $$\text{det} A:U \rightarrow \mathbb{R}^\times$$ is continuous.

Here's the kicker, though: Tu says that because of the Intermediate Value Theorem, $$(\text{det} A)(p)$$ is always positive or always negative on $$U$$, being unable to change signs without smoothly crossing over zero---thus showing that $$f$$ is constant on U. However, I have no idea what the intermediate value theorem means when not talking about functions from some closed interval of $$\mathbb{R}$$ to $$\mathbb{R}$$. Naturally, I'd say $$\text{det} A$$ being continuous means that $$\text{det} A \circ \phi^{-1}:\phi(U) \rightarrow \mathbb{R}$$ is continuous for a chart $$(U,\phi)$$. But, how is the intermediate value theorem extended to a function $$\mathbb{R}^n \rightarrow \mathbb{R}$$? My intuition is that, like in one dimension, the connectedness of $$U$$ is key to an analogous theorem, but can a generalization of the IVT be made without introducing notions of curves along $$U$$ and other differential geometry machinery into the mix? Can I apply the 1-D IVT to all possible curves between two points in $$U$$ or something? I get the intuitive sense of what he means by it, but how can I put it on more mathematical footing?

The determinant is a continuous function from a connected open set to $$\mathbb{R} -\{0\}$$, so its image is a connected set.
• And $\mathbb{R} - \{0 \}$ has two connected components, so the image can only be one of those, I see. Thanks for humoring me! Commented Dec 20, 2020 at 23:17
Usually the IVT is understood as a theorem of elementary calculus which deals with functions $$f : [a,b] \to \mathbb R$$. However, there is a generalized version.
Note that if $$f : X \to Y$$ is a continuous map and $$X$$ is connected, then $$f(X)$$ is connected. If $$Y = \mathbb R$$, then the connected subsets of it are precisely the intervals. Therefore
Let $$f : X \to \mathbb R$$ be a continuous map and $$X$$ be connected. If $$y_1 = f(x_1) < y_2 = f(x_2)$$, then for each $$y \in (y_1,y_2)$$ there exists $$x \in X$$ such that $$f(x) = y$$.
• I see. I suppose it is enough that $U$ is connected, so $\text{det} A$ maps to one of the connected halves of the separation $\mathbb{R}^{\times} = (- \infty, 0) \cup (0, \infty)$, but it is possible to specialize the result to something like an IVT. I suppose I just got hung up on the specific usage of IVT, which wasn't really necessary to even state in the proof. Thanks much! Commented Dec 20, 2020 at 23:15