Let $M$ be a simply connected smooth manifold ($M=R^3$ is enough for my purposes).
Then, the following Proposition 1 is obvious.
Proposition 1 For all piecewise smooth loop, $c[0,1]\to M$, there is $H$ such that;
(1) $H:I\to M$ is a continuous mapping, Here $I:=[0,1]\times[0,1]$
(2) For all t,s ($0\le t\le 1$, $0\le t\le 1$)
$H(t,0) =c(t)$
$H(0,s) = H(1,s)$
$H(t,1) =c(0)$
∵It is just a paraphrase of the definition of a simply connected space.■
And if there is piecewise smooth $H$, then, for all smooth one-form $\omega$ ($\omega$ is a differential 1-form on M), we can define $H^{*}\omega$ (Here, $H^{*}\omega$ is a pullback of the $\omega$ ) and when we integrate $d{H}^{*}\omega$ over the interval $I:=[0,1]\times[0,1]$ and integrate $H^{*}\omega$ over the $\partial I$, we can obtain following proposition:
Proposition 2
Let $\omega$ be a smooth One-form on $M$, $d\omega = 0$, and, $c:[0,1]\to M$ be a piecewise smooth loop on M, then,
${\oint}_{c} \omega = 0$
However, it is not obvious to me that, Can it happen that "there is the only $H$ that is not piecewise smooth but continuous ?"
My Doubt:
Whether the following is correct ?:
For all piecewise smooth loop, $c[0,1]\to M$, there is $H$ such that;
(1) $H:I\to M$ is a piecewise smooth mapping, (Here $I:=[0,1]\times[0,1]$ )
(2) For all t,s ($0\le t\le 1$, $0\le t\le 1$)
$H(t,0) =c(t)$
$H(0,s) = H(1,s)$
$H(t,1) =c(0)$
If correct, please give proof or site reference literature. If wrong, please tell me a counterexample and tell me "how to justify the proposition 2.""
Edit: Maybe one can prove it using this book ... --Added on September 25, 2019.