# path in a manifold

• Feb 4th 2011, 05:18 AM
Sogan
path in a manifold
Hello,

we have the following setting:
c:[0,1]->M a continuous path in a manifold M. I want to show that there has to be a finite open cover of coord. domains \$\displaystyle (U_i)_{1,..,n}\$ of c[0,1] , s.t. \$\displaystyle U_i \cap c[0,1]=c(a_i,b_i) \$ for ome \$\displaystyle a_i, b_i.\$

For sure, the image c[0,1] is compact in M. Therefore we can find a finite subcover to each open cover.
But why is the intercection of the finite cover with the curve of that form?

I think, we must first of all take open sets U_i, which has a connected intercection with c[0,1].
Why do we have such a cover?

I hope you can help me.

Regards
• Feb 4th 2011, 05:52 AM
HallsofIvy
Start by taking an open cover of [0, 1] with that form- that is, such that the intersection of each \$\displaystyle U_i\$ with [0,1] is an interval. Then look at c(U) for all U in the open cover of [0, 1]
• Feb 4th 2011, 05:56 AM
Bongo
I think there are some problems with your solution Hallsof Ivy!
First the U_i has to be coordinate domains and on the other hand the sets c(U) are not open in general!

But every Manifold is locally eucledean. Therefore we can shrink every open set, s.t. they are coordinate balls. Furthermore we know that there has to be an open cover of c[0,1] with such coordinate balls, s.t. every intersection is "nice".
• Feb 4th 2011, 06:12 AM
Sogan
Yes it is intuitively right, but why are the intersections so "nice" as you said?