This is quite a detailed question: I'm struggling to understand a few parts of a proof of the following Lemma. I've placed stars ($\bigstar$) where I'd like to draw your attention.

Lemma: Let $P:Y\to X$ be a covering projection and $f: Z \to X$ a continuous map where $Z$ is a simply connected and locally path connected space. Suppose given base points $x_0, y_0, z_0$ of $X,Y,Z$ with $p(y_0) = x_0 = f(z_0)$. Then there is a unique continuous $\tilde{f} : Z \to Y$ with $p \tilde{f} = f$ and $\tilde{f}(z_0) = y_0$.

The proof, summarised/paraphrased, is as follows:

Given $z \in Z$, choose a path $u$ from $z_0$ to $z$ in $Z$. Let $\tilde{u}$ be the unique lifting of $fu$ to a path in $Y$ starting at $y_0$, and define $\tilde{f}(z) = \tilde{u}(1)$.

Can then show $\tilde{f}$ is well-defined, and that it's the only possible mapping that could work. It remains to be shown that it is continuous.

To do so, let $z \in Z$ and let $V$ be an open neighbourhood of $\tilde{f}(z)$. Without loss of generality ($\bigstar$), we can assume $V$ is of the form $h^{-1}(U \times {d})$, where $U$ us an evenly covered neighbourhood of $p\tilde{f}(z) = f(z)$ and $h: p^{-1}(U) \to U \times D$ is a homeomorphism with $D$ discrete.

$f^{-1}(U)$ is an open neighbourhood of $z$, so contains a path connected neighbourhood $W$. For any $z' \in W$, we can choose a path from $z_0$ to $z'$ of the form $u.v$ where $u$ is a path from $z_0$ to $z$ and $v$ takes values in $W$. Now $fv$ takes values in $U$, which is evenly covered by $p$. So its lifting to a path in $Y$ starting at $\tilde{f}(z)$ must take values in $V \ (\bigstar \bigstar)$, and in particular $\tilde{f}(z') = \tilde{(u.v)}(1) \in V$. So $W \subseteq \tilde{f}^{-1}(V)$ and hence $\tilde{f}$ is continuous.

My confusions are as follows:

$\bigstar$: Is the following reasoning as to why this does not lose generality correct? Given any $y = \tilde{f}(z)$, consider $p(y)$. We can find an evenly covered neighbourhood $U$ containing $p(y)$. Then there is a homeomorphism $h : p^{-1} (U) \to U \times D$ where $D$ is a discrete space. Then $(p(y), d) \in U \times \{d\}$ for any $d \in D$ and $h^{-1} (p(y),d) \subseteq h^{-1} (U \times \{d\})$. But $h^{-1}(p(y),d) = p^{-1} p(y)$ which contains $y$.

$\bigstar \bigstar$: I'm confused with this bit. Intuitively, $p^{-1}(U)$ 'looks like' lots of copies of $U$, and $V$ is one of those copies. But I can't turn this into a definitive reason why $\tilde{v}$ (i.e. the lifting of $fv$) must be entirely contained within $V$. On a different note, am I correct in saying $V \cong U$? I'm convinced there's a simple set-theoretic explanation, but it is eluding me.

Any help will be greatly appreciated - I understand that this is a long post.

Thank you.

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+1 for the details and pointing out your question clearly. :) –  Srivatsan Nov 2 '11 at 14:18

The explanation you give for $\bigstar$ seems correct to me.

As for $\bigstar \bigstar$:
Since the trace of $\tilde v$ (that is the set of points $\tilde v ([0,1] )$ in $Y$ through which the path $\tilde v$ runs) is a connected set and since it is contained in the disjoint union of the open sets $h^{-1} (U\times\lbrace \delta \rbrace) \; (\delta \in D)$ it must be contained in exactly one of them, namely $V=h^{-1} (U\times\lbrace d \rbrace)$ since this is the one that contains $\tilde v(0)=\tilde f(z)$.

Finally, you are absolutely correct that $U$ and $V$ are homeomorphic.
The natural homeomorphism is the restriction $p|V:V\to U$, whose inverse is the continuous map $h^{-1}\circ j:U\to V$, where $j:U\to U\times D: u\mapsto (u,d)$ is the insertion of $U$ in the relevant slice of the product.

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$\bigstar$: Yes, that's fine. Covering projections can seem a strange concept at first, but I find that it helps a lot to bear in mind the typical example $p : \mathbb{R} \to S^1$ with $p(t) = e^{2 \pi i t}$, which I'm sure you'll have seen (imagine bending the real line into a helix, which projects down to cover the unit circle: our discrete space is just $\mathbb{Z}$). Given any point on the helix, it's clear you can find an open interval around that point which is one of the 'copies' of the interval on $S^1$ that it projects down onto under $p$.
$\bigstar \bigstar$: You have that $fv(t) \in U$ for all $t$. This means that $(fv(t),d) \in U \times \{d\}$ for all $t$. Now, you should know that by the definition of 'evenly covered', you have that $ph^{-1} (x,d) = x$ for all $x \in X$ and $d \in D$ (see here: Query about proof of homotopy lifting property). This means that $ph^{-1}(fv(t),d) = fv(t)$ for all $t$. But, by uniqueness of lifting, this exactly gives us the lifting of $fv$: we must have that $\tilde{v} = h^{-1}(fv(t),d)$. By the form of $V$, we have that $\tilde{v}(t) \in V$ for all $t$.