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