# 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

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In fact, when the theorem says $p\hat{u} = u$, does that mean $p\hat{u}(t) = u(t) \ \forall t$? If so, then I can see why. I was taking $p\hat{u} = u$ to mean that their images are the same. – Martin P Nov 1 '11 at 19:19
Because $p(h^{-1}(x,d))=x$. That's the nature of the homeomorphism between $h:p^{-1}(U)\rightarrow X\times D$. – Thomas Andrews Nov 1 '11 at 19:28

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 $p(h^{-1}(x,d)) = \pi_1 h h^{-1}(x,d) = \pi_1 (x,d) = x$.

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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 $h(\hat{u}(t)) = (u(t),d)$

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