Let $n \in \mathbb{N}$, $M$ be a set and let $\mathcal{A} = \{(\varphi_a, U_a)\}_{a \in \mathcal{A}}$ be a system of tuples so that:
$U_{a} \subseteq \mathbb{R}^n$ is open for all $a$;
$\varphi_a: U_a \to M$ is injective for all $a$, with image denoted $V_a$;
$\varphi_a^{-1} \circ \varphi_b: \varphi_b^{-1}(V_a \cap V_b) \to \varphi_a^{-1}(V_a \cap V_b)$ is a homeomorphism for all $(a, b)$.
Question: Can we in general conclude that there is a topology on $M$ so that $V_a$ is open in $M$ and $\varphi_a: U_a \to V_a$ is a homeomorphism for all $a$?
My attempts: I wanted to define $U$ to be open in $M$ if and only if $\varphi_a^{-1}(U \cap V_a)$ is open in $U_a$ for all $a$. Clearly this is a topology and also every $\varphi_a$ is continuous (regardless if we see it as a map to $V_a$ or M). But I have problems with the rest.
For example: Why is $V_a$ open in $M$? Is there anything which tells us that $\varphi_b^{-1}(V_a \cap V_b)$ is open in $U_b$ for all $b$?
If the statement is not true, then you could help me finding conditions that make it true. The question arose while trying to solve an exercise which - roughly speaking - says that the tangent bundle $TM$ is a smooth manifold when $M$ is one.
Thanks for any help.
The answer to your question is no. For example, let $X$ be the quotient of $\mathbb{R}\times\{0,1\}$ obtained by identifying $(q,0)$ and $(q,1)$ for each $q\in\mathbb{Q}$, and let $\mathcal{A} \;=\; \{(\mathbb{R},\varphi_0),(\mathbb{R},\varphi_1)\}$, where $\varphi_0$ and $\varphi_1$ are the two obvious inclusions of $\mathbb{R}$. Then $\varphi_0^{-1}\circ \varphi_1$ is the identity map on $\mathbb{Q}$, so it satisfies the conditions you've given.
However, there's no topology on $X$ that makes $\varphi_0$ and $\varphi_1$ into homeomorphisms with open image, since $\varphi_0^{-1}(V_0\cap V_1) = \mathbb{Q}$ isn't open in $\mathbb{R}$.
Intuitively, what's going on is that you need to place some restriction on how the $U$'s are glued together if you want to get a manifold. It should suffice to require that $\varphi_a^{-1}(V_a\cap V_b)$ is open in $U_a$ for each $a$ and $b$.