# if $V_1\cong U_1, V_2\cong U_2$, is $(V_1\cup V_2 \cong U_1\cup U_2)$? Pasting homeomorphisms

My question arises from the theory of covering spaces. assume $f:Y\to X$ is a covering map, or more generally a local homeomorphism. Assum $U_1,U_2\subset X$ are open sets such that $f|_{V_1}, f|_{V_2}$ are homeomorphism onto $U_1, U_2$ respectively. assume $V_1\cap V_2 \not=\emptyset$.

Is it true that $f|_{V_1\cup V_2}$ is a homeomorphism onto $U_1 \cup U_2$?

My guess is yes, as it is very reasonable geomtricaly. Moreover, the only problem could be with the injectiveness, that should probably be okay because we made $V_1\cap V_2 \not=\emptyset$ so we fixed a "gluing" point between them. I can see that it is enough to show that for $y_1\in V_1-V_2, y_2\in V_2-V_1$ we have $f(y_1)\not=f(y_2)$, but I can't manage to show that.For that it would suffice to show $f(V_1-(V_1\cap V_2))=U_1-U_2$ but then again I can't prove it.

Take $X=Y=S^1$ (with elements angles $\operatorname{mod} 2\pi$) and $f\colon \theta\mapsto 2\theta$ the usual $2$-fold covering. Let $V_1=(0,\pi)$ and $V_2=(\frac{\pi}{2},\frac{3\pi}{2})$ such that $V_1\cap V_2=(\frac{\pi}{2},\pi)\neq\varnothing$. Now $f|_{V_1}$ and $f|_{V_2}$ are both homeomorphisms onto their images $U_1=(0,2\pi)$, $U_2=(\pi,3\pi)$, respectively. However, $V_1\cup V_2=(0,\frac{3\pi}{2})$ is mapped to all of $S^1$ by $f$, so $f|_{V_1\cup V_2}$ is not a homeomorphism onto its image!