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DISLAIMER: I probably misunderstood the statement I'm referring to in this question, so the question refers to a wrong statement. My answer below provides a (hopefully proper) 'fix' of this misunderstanding. I'm not sure this will be of help to others, but who knows.

Let

  • $p:\,E \longrightarrow B$ be a covering, i.e. a locally trivial map with discrete fibers
  • $E$ be locally path connected
  • $Aut(p)$ be the set of topological automorphisms $\alpha$ over $E$ such that $p\alpha=p$ and $H \lt Aut(p)$ be a subgroup
  • $U$ be open, path-connected and evenly covered by $p$
  • $\mathop{can}: E \longrightarrow E/H$ be the canonical projection onto the orbit space

In Dieck's "Algebraic Topology", page 65, it says that for all sheets $U_i\subset E$ (summands of the disjoint union $p^{-1}(U) = \bigcup_i U_i$ in $E$) the restriction \begin{align*} q|: \bigcup_i \mathop{can}(U_i) \longrightarrow U \end{align*} of the induced map \begin{align*} q: E/H \longrightarrow B \end{align*} by which $p$ factors through the orbit space is bijective. Surjectivity is obvious, but how can I show that this map is injective?

My thoughts so far:
Let $$ p(x) = q([x]) = q([x']) = p(x') $$ for $x,x' \in \bigcup_i U_i$. Then $x, x' \in p^{-1}(\{p(x)\})$.

So I need to find some $h \in H$ such that $h(x) = x'$ ($\iff [x] = [x']$, which I want to show). However I'm not sure how to find this. I think that I have to use what Dieck mentions in the same paragraph:

  • Any $h \in Aut(p)$ permutes the sheets, since they are the path components of $p^{-1}(U)$,

together with the fact that

  • Each such $h$ also restricts to an automorphism over $p^{-1}(\{p(x)\})=p^{-1}(\{p(x')\})$.

But I'm not sure/not really making much progress right now...

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It seems like I understood the wording in the book (which was somewhat scarce). What got me confused that we have homeomorphisms

$$ p|_{U_i}: U_i \longrightarrow U $$

and the wrong idea that

$$ \begin{equation}\label{eq}\tag{I} \mathop{can}(U_i) = U_i/H \end{equation} $$ so that a homeomorphy $$ q|_{can(U_i)}: can(U_i) \longrightarrow U $$ seemed unreasonable, since canonical projections like $U_i \longrightarrow U_i/H$ generally make the set 'smaller'.

However, equations such as $\eqref{eq}$ don't generally hold true (a set-equality would hold if there were only one sheet; I'm not sure one could generally say this would be a homeomorphy, though).
In the case of several sheets, we have on the left side of $\eqref{eq}$ classes that have representatives in $\bigcup_i U_i$, on the right side they are from $U_i$ for a single $i$ of the index set.

So here goes my answer with a hopefully correct re-interpretation of the question/text:

$$ \forall j \bigl(\, q|:\, \mathop{can}(U_j) \longrightarrow U \bigr) $$ is a bijection.

Proof: Trivial; assume $$ p(x)=q([x])=q([x'])=p(x') $$ for $[x],\,[x'] \in \mathop{can}(U_j)$. By picking representatives, can w.l.o.g. assume $x,\,x' \in U_j$. But then, $x=x'$ since $p$ is a homoemorphism over each sheet$._{\;\square}$

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    $\begingroup$ Nitpick: It's kind of pointless to say "trivial" in a proof. If it is indeed trivial, then your proof will be short and clear. If it is not trivial, then you should not say that it is. :) $\endgroup$ – Neal Jun 30 '17 at 18:37
  • $\begingroup$ You have a point :P $\endgroup$ – polynomial_donut Jun 30 '17 at 18:40
  • $\begingroup$ @Neal I was almost going to remove the word 'Trivial', but doubts of doing so occured: If it was indeed better to remove it, then the text would have been short and clear without it. However, removing the word does not make a significant difference to the text, so I won't remove it. $\endgroup$ – polynomial_donut Jun 30 '17 at 18:50

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