Hello I have the following theorem where I have problems with one step in the proof:
Let $W$ be locally connected and $p:X \to Y$ a covering. Let $F:W\times I \to Y$ be a homotopoy and $f:W \times \{0\} \to X$ a lifting of the restriction of $F$. Then there is a unique homotopy $G:W\times I \to X$ making everything comutes.
The proof goes (in my interpretation, is this right?) as following: For every $w\in W$ we have a path $f_w:I\to Y$ given by $t\mapsto F(w,t)$. We note that $f_w(0)= F(w,0) = p(x_0)$ for some $x_0 \in Y$ (has to exist as p is surjective) chosen s.t $x_0 = f(w,0)$. Then by the lifting property there is a unique path $g_w:I\to X$ sucht that: $pg_w = f_w$ and $g_w(0) =x_0$. And we get a lifting by defining $G(w,t) = g_w(t)$. Now $G$ comutes with everything (we builded everything in sucht a way that this works). The unique thing to proof is that $G$ is continuos.
Here my Problems are starting. My Professor now just used the elementary Sets $U_i$ covering $Y$ and said that for $x\in W$ by a previouse Lemma we find $N \subset W$ connected (as W is locally connected) neighborhood of $x$. and $n>0$ s.t $F(N\times [i/n, i+1/n])\subset U_i$ for some $i$. But now I dont understand what is happening (the Prof just said that $G$ is continuos on $N\times \{0\}$ as here it is $f$ and then Inductvly it is continuos on $N\times \{i/n\}$ for all $i$. Here I dont understand how the induction works, and why this would sufficies to show that $G$ is continuos.
The proof shows that $G$ is continuous on $N \times [0, i/n]$ by induction on $i$. In the induction step, it suffices by the pasting lemma to show that $G$ is continuous on $N \times [i/n, (i + 1)/n]$ and by the induction hypothesis we know that it is continuous on $N \times \{i/n\}$. Now by assumption $F(N \times [i/n, (i + 1)/n])$ is contained in some elementary set $U = U_j$ (not the same $i$, which may be a source of your confusion). So $G(N \times [i/n, (i + 1)/n])$ is contained in $\sqcup_k V_k$ such that each restriction $p :V_k \to U$ is a homeomorphism.
Since $N \times \{i/n\} \cong N$ is connected, and $G$ maps it continuously, $G(N \times \{i/n\})$ is contained in a single $V = V_k$. But for each $x \in N$, $G$ also maps $\{x\} \times [0,1]$, in particular $\{x\} \times [i/n, (i + 1)/n]$ continuously, so all of those must also land in $V$. But now on $N \times [i/n, (i + 1)/n]$, $G$ must be given by $s \circ F$, where $s : U \to V$ is the continuous inverse of $p|V$, which completes the induction step.