In case of a homotopy $h: I^2 \rightarrow S^1$ we can define lifting as such an $\tilde{h}: I^2 \rightarrow \mathbb{R}$ that $e^{i\tilde{h}}=h$. The existence of $\tilde{h}$ requires a proof. A way to prove this fact I'm thinking of, is to devide $I^2$ on such squares that on each of them $|h(x) - h(y)| < 2$, i.e. "angle" between any two values of $h$ differ less than on $\pi$.

Now, for the square containing 0 we define $\tilde{h}(z) := \Delta_z$, where $h(z) = h(0)e^{i\Delta_z}$. Of course I'd like to this for the second square, but I need tese two mappings to match on the common edge. How to achieve this?

In case of analogues theorem for $f:I \rightarrow S^1$ this problem, I think, doesn't arise as the two consequitive intervals have only a single point in common (so we can set $\Delta_1 + \Delta_z$ for the second interval for example, where $\Delta_1$ is the value of $\tilde{f}$ at the first intersection point.


Hint: remark that $I^2=I\times I =[0,1]\times [0,1]$. Show that you can lift $h(t,0)$ to $H_0:[0,1]\rightarrow R$ such that $p \circ H_0=h(0,t)$ where $p:R\rightarrow S^1$.

Then show that you can lift $h(t,0)$ by a map $H'_0:I\rightarrow R$ such that $H_0(0,0)=H'(0,0)$. For every $t$, there is a unique $H_t:t\times I$ such that $H_t(0)=H'(t,0)$ and $p(H_t(v))=h(t,v)$. The map $H:I^2\rightarrow R$ defined by $H(t,v)=H_t(v)$ is continue and lifts $h$.

  • $\begingroup$ Thank you! Slicing was my first idea towards this problem. One thing, may be I miss something, but is it really that obvious that $H(t,v)$ is continues? $\endgroup$ – Stan Feb 10 '16 at 0:10

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