1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.