Clarification needed on (one of) the definition(s) of differentiable manifold This is taken from page 3 of Olver’s Application of Lie Groups to Differential Equations:

I can’t figure out a proof for the highlighted statement. More precisely, what I need to prove is that the collection $$\bigcup_\alpha \left\{ \chi_\alpha^{-1}(W) \colon W \text{ is an open subset of } V_\alpha \right\}$$ satisfies the second condition for a basis (the following definition is taken from Munkres’ Topology):

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal B$ of subsets of $X$ (called basis elements) such that

*

*For each $x \in X$, there is at least one basis element $B$ containing $x$.


*If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$.

I thought it would be enough to show that, for any 4-tuple $(\alpha,W_1,\beta,W_2)$ such that $W_1$ is an open subset of $V_\alpha$ and $W_2$ is an open subset of $V_\beta$, having set $B_1 =\chi_\alpha^{-1}(W_1)$ and $B_2 = \chi_\beta^{-1}(W_2)$, the set $\chi_\alpha(B_1 \cap B_2)$ is open. In order to do so, I noticed that $$\chi_\alpha(B_1 \cap B_2) = W_1 \cap \left(\chi_\alpha \circ \chi_\beta^{-1}\right) \! \left(W_2 \cap \chi_\beta\!\left(U_\alpha \cap U_\beta\right)\right)$$

My problem is, how do I know that $\chi_\beta\!\left(U_\alpha \cap U_\beta\right)$ is open? This is also necessary for condition (b) of definition 1.1 to make sense, which makes me think that maybe it is an implicit assumption, but I’d like to be sure about that. Thanks for the help.
 A: I think it is implicitly an assumption on Olver's part that the sets $\chi_\alpha(U_\alpha\cap U_\beta)$, $\chi_\beta(U_\alpha\cap U_\beta)$ are open subsets of the Euclidean space. In the second paragraph on p. 4,

The degree of differentiability of the overlap functions $\chi_\beta\circ\chi_\alpha^{-1}$ determines the degree of smoothness of the manifold $M$. We will be primarily interested in smooth manifolds, in which the overlap functions are smooth, meaning $C^\infty$, diffeomorphisms on open subsets of $\mathbb R^m$ [emphasis added].

That being said, I do not know if the minimal assumptions (a), (b), (c) of Definition 1.1. actually might imply already that $V_\alpha := \chi_\alpha(U_\alpha\cap U_\beta)$, $V_\beta := \chi_\beta(U_\alpha\cap U_\beta)$ are both open subsets of $\mathbb R^m$. I think of the following Proposition as a sign of the potential nuance involved in proving $V_\alpha,V_\beta$ are open, if this can be proved at all.
Proposition. If at least one of $V_\alpha$, $V_\beta$ is open, then they are both open.
This Proposition seems like a no-brainer, and maybe even a little pedantic to state because it seems obvious that if $V_\alpha$ looks like a unit ball with $C^0$ glasses on, then $V_\beta$ also looks like a unit ball with $C^0$ glasses on. It is nontrivial however to actually prove that a homeomorphism won't send a nice open ball to some set with a "boundary". This problem is referred to as Topological Invariance of the Boundary. When we switch to $C^1$ glasses however, the smoothness assumption of the transition maps makes the rigorous proof of the Smooth Invariance of the Boundary Proposition simpler, by comparison. (The proof$^{[1]}$ of the $C^1$ case still invokes the $C^1$ inverse function theorem, a theorem that is often seen as the capstone of upper-level real analysis courses). I offer this Proposition as a way of stating I think that the issue of whether $V_\alpha,V_\beta$ are open is perhaps a comparatively delicate question compared to the material in the given context in this part of the book.
I underline this is nothing like a "proof" of the Vague Claim: It is hard to prove that $V_\alpha,V_\beta$ are open. I just don't know how to prove it.

[1]: See Theorem 1.46 in John Lee's book Introduction to Smooth Manifolds, 2nd Ed. for a proof of the Smooth Invariance of the Boundary and Problem 17-9 for the proof of the Topological Invariance of the Boundary using the concept of de Rham cohomology.
A: Take $x\in \chi_\alpha^{-1}(W_\alpha)\cap\chi_\beta^{-1}(W_\beta)$, where $W_\alpha$ is some open subset of $V_\alpha\subset \mathbb{R}^m$, the range of the $\alpha$-th chart (and analogously for the $\beta$ one).
Let $W'_\alpha=(\chi_\alpha\circ\chi_\beta^{-1})(W_\beta)$. It is an open neighbourhood of $\chi_\alpha(x)$ contained in $V_\alpha$.
Then $G:=W'_\alpha\cap W_\alpha$ is an open nbd of $\chi_\alpha(x)$, and you can check that this $G$ yields by preimage under $\chi_\alpha$ an open set $\chi_\alpha^{-1}(G)$ sitting within $\chi_\alpha^{-1}(W_\alpha)\cap \chi_\beta^{-1}(W_\beta)$, and therefore Condition 2 in your definition of basis is satisfied.
Remark: I am implicitly using that a finite intersection of open sets is open and that open sets map to open sets under homeomorphisms. Note $\chi_\alpha,\chi_\beta$ are indeed homeomorphisms.
