Showing the existence of a topology such that $ \{ (U_\alpha, \phi _\alpha )\}$ forms an atlas.

Question

The problem verbatim:

Suppose X is a set covered by a collection of subsets $\{ U_\alpha \}$. We are given one-to-one correspondence $\phi _\alpha : U_\alpha \to V_\alpha \subseteq \mathbb{R^n}$ (which I believe is open) such that the (transition maps) $\psi _{\alpha \beta} : \phi_\beta (U_\alpha \cap U_\beta) \to \phi_\alpha (U_\alpha \cap U_\beta) $ given by $ \psi_{\alpha \beta} = \phi_\alpha (\phi_\beta)^{-1}$ is a homeomorphism. Show that there is a natural topology on $X$ such that the $ \{ (U_\alpha, \phi _\alpha )\}$ forms an atlas for $X$.

The way I constructed this topology is by saying: A set $O \subseteq X$ is open if $\phi_\alpha (\phi_\beta(X \cap O))^{-1} $ is open in $\mathbb{R^n}$. Since this way, I'd make sure that every mapping from open set in $X$ to $\mathbb{R^n}$ would be a homeomorphism. But I don't think it's correct. Furthermore, I am lacking in intuition too. Help will be much appreciated.

Answer 1

In your construction it is not clear what $\phi_\alpha (\phi_\beta(X \cap O))^{-1}$ means. Since $O \subset X$, you have $X \cap O = O$ which does not make much sense as an input for $\phi_\beta$. Probably you mean $U_\beta \cap O$. Moreover, with $M = \phi_\beta(U_\beta \cap O)$, what is $\phi_\alpha(M)^{-1}$? Perhaps you mean $\phi_\alpha^{-1}(M)$?

Nevertheless I think that you had the correct idea.

We are given a family $\mathcal A = \{(U_\alpha,\phi_\alpha) \}$ such that

1. Each pair $(U_\alpha,\phi_\alpha)$ consists of a subset $U_\alpha \subset X$ and a one-to-one correspondence, i.e. a bijection, $\phi_\alpha : U_\alpha \to V_\alpha \subset \mathbb R^n$.

2. The $U_\alpha$ cover $X$.

We want to construct a natural topology on $X$ such that $X$ becomes a manifold with atlas $\mathcal A$. That is, for each $\alpha$, the pair $(U_\alpha,\phi_\alpha)$ has to form a chart on $X$. This means

(a) $V_\alpha$ is open in $\mathbb R^n$

(b) $U_\alpha$ is open in $X$

(b) $\phi : U_\alpha \to V_\alpha$ is a homeomorphism.

Thus, to have a chance to construct the desired topology on $X$, we have to make two basic assumptions on $\mathcal A$.

(B1) For each $\alpha$, $V_\alpha$ is open in $\mathbb R^n$.

(B2) For any two $\alpha, \beta$, $\phi_\alpha(U_\alpha \cap U_\beta)$ is open in $V_\alpha$.

Requirement (B1) is completely obvious. To see why (B2) is an indispensable requirement, note that that we want the sets $U_\alpha$ to be open in $X$ and the $\phi_\alpha$ to be homeomorphisms. Then also each intersection $U_\alpha \cap U_\beta$ will be open in $U_\alpha$ and the image $\phi_\alpha(U_\alpha \cap U_\beta)$ open in $V_\alpha$.

The desired topology on $X$ must obviously have the property that for any open $W \subset X$ we have

$(*) \quad$ $\phi_\alpha(W \cap U_\alpha)$ is open in $V_\alpha$ (or equivalently, open in $\mathbb R^n$) for all $\alpha$.

So let us consider the set $\mathcal T$ of all $W \subset X$ satisfying $(*)$. We claim that this is a topology on $X$. In fact, it is the finest topology $\mathcal T$ on $X$ such that $(*)$ is satisfied for all $W \in \mathcal T$.

Clearly $\emptyset, X \in \mathcal T$. If $W_\iota \in \mathcal T$, then $\phi_\alpha((\bigcup W_\iota) \cap U_\alpha) = \phi_\alpha(\bigcup (W_\iota \cap U_\alpha)) = \bigcup \phi_\alpha(W_\iota \cap U_\alpha)$ is open in $V_\alpha$, thus $\bigcup W_\iota \in \mathcal T$. If $W, W' \in \mathcal T$, then $\phi_\alpha((W \cap W') \cap U_\alpha) = \phi_\alpha((W \cap U_\alpha) \cap (W' \cap U_\alpha)) = \phi_\alpha(W \cap U_\alpha) \cap \phi_\alpha(W' \cap U_\alpha)$ is open in $V_\alpha$, thus $W \cap W' \in \mathcal T$.

We claim that

$(**) \quad$ Let $W \subset U_\beta$. Then $W \in \mathcal T$ if and only if $\phi_\beta(W)$ is open in $V_\beta$.

One half is trivial. The non-trivial part is to show that if $\phi_\beta(W)$ is open in $V_\beta$, then $\phi_\alpha(W \cap U_\alpha)$ is open in $V_\alpha$ for all $\alpha$. Since $\phi_\beta(W)$ is open in $V_\beta$, we see that $\phi_\beta(W \cap U_\alpha) = \phi_\beta(W \cap (U_\beta \cap U_\alpha)) = \phi_\beta(W) \cap \phi_\beta(U_\beta \cap U_\alpha)$ is open in $V_\beta \cap \phi_\beta(U_\beta \cap U_\alpha) = \phi_\beta(U_\beta \cap U_\alpha)$. Therefore $\phi_\alpha(W \cap U_\alpha) = \psi_{\alpha \beta}(\phi_\beta(W \cap U_\alpha))$ is open in $\phi_\alpha(U_\beta \cap U_\alpha)$ and thus open in $V_\alpha$. Here we have to invoke (B2).

But $(**)$ proves that each $\phi_\beta : U_\beta \to V_\beta$ is a homeomorphism. In other words, $\mathcal A$ forms an atlas for $X$.