Use parallel transport to explain the homotopy lifting property

Question

Suppose $p:E\rightarrow B$ is a fiber bundle, Hatcher's algebraic topology proposition 4.48 proves that for any CW complex $X$, we have the homotopy lifting property of $X$ with respect to $p$. Meaning, if $F:X\times[0,1]\rightarrow B$ is a homotopy, and $g:X\rightarrow E$ is a map satisfying $p\circ g(x)=F(x,0)$, then $g$ can be extended to a homotopy $G:X\times[0,1]\rightarrow E$ such that $p\circ G=F$.

Now if in addition $p$ is a principle $G$ bundle for some Lie group $G$, then my understanding is that the above homotopy lifting property can be understood in terms of parallel transport with respect to the Ehresmann connection of the bundle. But I can't seem to find a reference for this. Can anyone explain this to me or point out a reference?

Actually, do we just lift each curve $t\rightarrow F(x,t)$? How do we show in this way we get a smooth homotopy $G$?

Answer 1

I only know the following:

If you're just lifting a path $\gamma:[0,1]\to B$ up to $P$, then you can always get a horizontal lift $\hat\gamma$ starting at any point $p_0\in P$ above the start of the path, (equivalently, can parallel transport of $p_0$ along $\gamma$).

This is done by:

1. Start with any (smooth) lift $\tilde\gamma$ of $\gamma$.
2. Write out the "horizontal condition" equation $$\omega\big(\frac{\mathrm d}{\mathrm d t}[\tilde\gamma(t).g(t)] \big)=0$$ as a differential equation for $g(t)$, a path in $G$ of gauge transformations which gauge transforms this to a horizontal path. This turns out to be a first order ODE, so we can solve it, with initial conditions, smoothness, etc.

Then $\hat\gamma(t) := \tilde\gamma(t).g(t)$ is your horizontal lift. This is in Kobayashi+Nomizu.

But for a more general $X$, I think if you try to do the same thing, the horizontal condition will be a PDE. It seems like you could come up connections for which there is no horizontal lift of a particular submanifold $X$.