In Sect. 1.4 "Global Continuation of Functions on Simply-Connected Manifolds" of Jost's book Compact Riemann Surfaces, a lemma is exhibited there. And I am having trouble understanding part of its proof.
Lemma 1.4.1 Let $M$ be a simply connected manifold, and $\{U_\alpha\}$ an open covering of $M$, assume that all the $U_\alpha$ are connected. Suppose given on each $U_\alpha$ a family $F_\alpha$ of functions (not satisfying $F_\alpha=\varnothing$ for all $\alpha$) with the following properties:
i) if $f_\alpha\in F_\alpha,\ f_\beta\in F_\beta$ and $V_{\alpha\beta}$ is a component of $U_\alpha\cap U_\beta$, then $$f_\alpha\equiv f_\beta\text{ in a neighborhood of some }p\in V_{\alpha\beta}$$ implies $$f_\alpha\equiv f_\beta\text{ on }V_{\alpha\beta}$$
ii) if $f_\alpha\in F_\alpha$ and $V_{\alpha\beta}$ is a component of $U_\alpha\cap U_\beta$, then there exists a function $f_\beta\in F_\beta$ with $$f_\alpha\equiv f_\beta\text{ on }V_{\alpha\beta}$$
Then there exists a function on $M$ such that $f|_{U_\alpha}\in F_\alpha$ for all $\alpha$. Indeed, given $f_{\alpha_0}\in F_{\alpha_0}$, there exists a unique such $f$ with $f|_{U_{\alpha_0}}=f$.
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Proof. Consider the set of all pairs $(p,f)$ with $p\in U_\alpha,\ f\in F_\alpha$ ($\alpha$ varies over all possible indices). Call two pairs $(p,f)$ and $(q,g)$ equivalent if $p=q$ and $f=g$ in some neghborhood of $p$. Let $[p,f]$ be the equivalence class of $(p,f)$ and $M^*$ the set of all equivalence classes. Define $\pi:M^*\to M$ by $\pi([p,f])=p$. For $f_\alpha\in F_\alpha$, let $U'(\alpha,f_\alpha)=\{[p,f_\alpha]:p\in U_\alpha\}$... (omitted, not related to my question).
The rest of the proof involves defining a topology on $M^*$. The book does it as follows:
For a subset $\Omega\subset M^*$, if $\Omega\subset U'(\alpha,f_\alpha)$ for some $U'(\alpha,f_\alpha)$, then it is called open if $\pi(\Omega)\subset U_\alpha$ is open; if $\Omega$ is arbitrary, then it is called open if $\Omega\cap U'(\alpha,f_\alpha)$ is open for all $\alpha$.
My question: Since $U_\alpha$ is open by assumption, $\pi(\Omega)$ is open in $U_\alpha$ is equivalent to being open in $M$. So why bother defining the topology using $U'(\alpha,f_\alpha)$, why not just say $\Omega$ is open if $\pi(\Omega)$ is open?
The issue is that $\pi(\Omega)\cap U_\alpha$ might be larger than $\pi(\Omega\cap U'(\alpha,f_\alpha))$, since there might be points of $\Omega$ that have the same first coordinate as points in $U'(\alpha,f_\alpha)$ but which are not in $U'(\alpha,f_\alpha)$. So just knowing that $\pi(\Omega)$ is open does not necessarily tell us that $\pi(\Omega\cap U'(\alpha,f_\alpha))$ is always open; the latter condition is stronger. The point is to make $\pi$ a local homeomorphism when restricted to each $U'(\alpha,f_\alpha)$, and this would not be guaranteed by your global definition.
(In fact, it is not even clear that your definition defines a topology: if $\pi(\Omega)$ and $\pi(\Omega')$ are open, why must $\pi(\Omega\cap \Omega')$ be open? This is not an issue when restricted to a single $U'(\alpha,f_\alpha)$, since then $\pi$ is injective.)