@Balarka The map belonging to $\varphi_{\beta}^{-1}\varphi_{\alpha}$ should be $\phi_{\beta\alpha}$ (not $\phi_{\alpha\beta}$), no? Then, $\mathrm{id}=\varphi_{\alpha}^{-1}\varphi_{\beta}\varphi_{\beta}^{-1}\varphi_{\gamma}\varphi_{\gamma}^{-1}\varphi_{\alpha}$ (on $U_{\alpha\beta\gamma}$), so $I=\phi_{\alpha\beta}\phi_{\beta\gamma}\phi_{\gamma\alpha}$; I don't see the identity working out otherwise. Now, consider the map $\coprod_{\alpha}U_{\alpha}\times\mathbb{R}^k\rightarrow E$ given by $(\alpha,(x,v))\mapsto\varphi_{\alpha}(x,v)$. This map is continuous on each summand, cause $\varphi_{…

Ok, let $f\colon X\rightarrow Y$ be a continuous, proper bijection and $Y$ a first-countable Hausdorff space. Let $A\subseteq X$ be closed. Consider a sequence $(y_n)_n$ in $A$ converging to a $y\in Y$. Qua definition, write $y_n=f(x_n)$ for some $x_n\in A$ for each $n$, and by surjectivity, write $y=f(x)$ for some $x\in X$. By convergence, $\{f(x),f(x_1),\dotsc,f(x_n),\dotsc\}$ is a compact set (one element of an open cover contains $f(x)$, hence all but finitely many elements of the sequence). Properness and bijectivity yield that $\{x,x_1,\dotsc,x_n,\dotsc\}$ is a compact set. Hence, it …