Proof:

The idea behind the proof is simple and parallels the proof of the previous theorem except that the book-keeping gets a bit more involved. Consider a covering ${\cal O}$ of $X$ by evenly covered open neighborhoods and choose a Lebesgue number $\epsilon$ for the covering

$$\{F^{-1}(U)/ U\in {\cal O}\}. \eqno(16.3)$$

Choose $n$ so large that any square in $[0, 1]\times [0, 1]$ of side $1/n$ is contained in one of the sets $F^{-1}(U)$ in (16.3). Partition $[0, 1]\times [0, 1]$ using the grid points

$$\Big\{\Big(\frac{j}{n}, \frac{k}{n}\Big)/ 0\leq j \leq n,\; 0\leq k \leq n\Big\}$$

and $S_{j, k}$ be the square with vertices

$$\Big(\frac{j}{n}, \frac{k}{n}\Big),\; \Big(\frac{j+1}{n}, \frac{k... ...ig(\frac{j+1}{n}, \frac{k+1}{n}\Big),\; \Big(\frac{j}{n}, \frac{k+1}{n}\Big).$$

Figure: Homotopy lifting property

[width=.4]GKSBook/fig13/fig13.eps

Let $U_{0, 0}$ be an evenly covered neighborhood in $X$ such that $F(S_{0, 0})\subset U_{0, 0}$ and ${\tilde U}_{0, 0}$ be the sheet in ${\tilde X}$ lying above $U_{0, 0}$. Denoting by $p_{0, 0}$ and $F_{0, 0}$ the restrictions of $p$ and $F$ to ${\tilde U}_{0, 0}$ and $S_{0, 0}$ respectively, define

$${\tilde F}_{0, 0} = p_{0, 0}^{-1} \circ F.$$

Thus ${\tilde F}_{0, 0}: S_{0, 0} \longrightarrow {\tilde X}$ is continuous, takes the value ${\tilde x}_0$ at the origin and is a part of the lift ${\tilde F}$ under construction. As in the previous theorem we shall construct the lift ${\tilde F}$ piece by piece and we now turn to the adjacent square $S_{1, 0}$ which is mapped by $F$ to an evenly covered neighborhood $U_{1, 0}$ in the cover ${\cal O}$. In particular (referring to the figure) $F(B) \in U_{1,0}$. Choose a sheet ${\tilde U}_{1,0}$ lying above $U_{1, 0}$ containing ${\tilde F}(B)$ and the restriction

$$p_{1,0} = p\Big\vert _{{\tilde U}_{1,0}}$$

maps ${\tilde U}_{1,0}$ homeomorphically onto $U_{1, 0}$. Now we define the next piece of the lift ${\tilde F}_{1,0}$ as

$${\tilde F}_{1,0} = p_{1,0}^{-1}\circ F$$

which is continuous on the square $S_{1, 0}$ and

$$p\circ {\tilde F}_{1,0} = F\Big\vert _{S_{1,0}}$$

In order to glue together the pieces ${\tilde F}_{0,0}$ and ${\tilde F}_{1,0}$ we must ensure that they agree all along the common edge $BC$ of the adjacent squares $S_{0, 0}$ and $S_{1, 0}$. Their restrictions along $BC$ where $t = 0$ and $0 \leq s \leq 1/n$ agree at $B$ namely

$${\tilde F}_{0,0}(0, \frac{1}{n}) = {\tilde F}_{1,0}(0, \frac{1}{n})$$

and are both lifts of the map

$$s\mapsto F(s, \frac{1}{n}),\quad 0\leq s \leq \frac{1}{n}$$

which implies, by uniqueness of lifts,

$${\tilde F}_{0,0}(s, \frac{1}{n}) = {\tilde F}_{1,0}(s, \frac{1}{n}),\quad 0 \leq s\leq \frac{1}{n},$$

as desired. It is now clear how the construction ought to proceed and we get a lift ${\tilde F}: [0, 1]\times [0, 1]\longrightarrow {\tilde X}$ of $F$.

We now have to check that ${\tilde F}$ is indeed a homotopy of paths with fixed endpoints. Well,

$$p\circ {\tilde F}(s, 0) = F(s, 0) = x_0,$$ for all $$s \in [0,1]$$

so that the connected set

$$\{{\tilde F}(s, 0)/ 0\leq s \leq 1\}$$

is contained in the discrete set $p^{-1}(x_0)$ and so must reduce to a singleton. Likewise ${\tilde F}(s, 1)$ is constant as $s$ varies over $[0, 1]$. Also $p\circ {\tilde F}(0, t) = F(0, t) = \gamma_1(t)$ and $p\circ {\tilde F}(1, t) = F(1, t) = \gamma_2(t)$ showing that ${\tilde F}$ is the desired homotopy between the lifts of $\gamma_1$ and $\gamma_2$ starting at ${\tilde x}_0$. $\square$