Query about proof of homotopy lifting property Let $p : Y \to X $ be a covering projection. Given a path $ u : I \to X $ and a point $ y \in Y $ with $p(y) = u(0) $, there exists a unique path $ \hat{u}:I \to Y $ with $p \hat{u} = u$ and $ \hat{u}(0) = y$. 

The proof I have of this proceeds as follows:
Consider first the case where the image of $u$ is contained within an open set $U \subseteq X$  that's evenly covered by $p$. Let $h: p^{-1}(U) \to U \times D$ be a homeomorphism, where $D$ is a discrete space. Then we can write $ h(y) = (p(y),d) = (u(0),d) $ for some $d \in D$. If $\hat{u}$ is any lifting of $u$, then the composite $ \pi_2 h \hat{u} : I \to p^{-1}(U) \to U \times D \to D $ must be constant (it's a continuous map from a connected space to a discrete space). If in addition $ \hat{u}(0) = y$, then this must be the constant map with value $d$. 
So the only possibility for  $\hat{u}$ is the mapping $ t \mapsto h^{-1}(u(t),d)$.

I understand everything up to this last sentence. Why "$u(t)$"? Why not $u(g(t))$, where $g$ is some continuous function with $g(0) = 0$? It's probably something simple, but I can't see it. Any help would be greatly appreciated.
Thanks
 A: The proof fails to state one last condition on $h$, namely, that $p$ on $p^{-1}(U)$ factors through $h$ - that is, the composition map $p^{-1}(U)\rightarrow U\times D \rightarrow U$ is $p_{|p^{-1}(U)}$, where $U\times D \rightarrow U$ is the natural projection.
So, $p(h^{-1}(x,d))=x$ for all $x\in U$.
So, if $p(\hat{u}(t)) = u(t)$ for all $t$, then $h(\hat{u}(t)) = (u'(t),d(t))$, for some $d:I\rightarrow D$.  Since $D$ is discrete, though, $d(t)$ must be constant.
On the other hand, $u(t)=p(\hat{u}(t)) = p(h^{-1}(h(\hat{u}(t)))) = p(h^{-1}(u'(t),d)) = u'(t)$
So, $u'(t)=u(t)$, and $h(\hat{u}(t)) = (u(t),d)$
A: To answer your comment: yes, $ p \hat{u} = u$ is an expression of equality as functions i.e. $ p\hat{u}(t) = u(t) $ for all $t$. 
I'll expand a bit on Thomas Andrews' comment. Suppose $ p :Y \to X$ is a continuous map. Then a subset $U$ of $X$ is evenly covered by $p$ if there's a homeomorphism $ h: p^{-1}(U) \to U \times D$, where $D$ is a discrete space, such that $\pi_1 h = p$ (where $\pi_1$ is projection onto the first factor). So $ p(h^{-1}(x,d)) = \pi_1 h h^{-1}(x,d) = \pi_1 (x,d) = x$. 
