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I'm self learning Rotman's Algebraic Topology and I've come across the proof in the below picture.

I don't understand how the proof shows that $S=p_*\pi_1(\tilde X, \tilde x_1)$. The proof initially starts with conjugate groups and then uses the commutative diagram to show the equality. But I don't understand what the link is here between the graphs being conjugate and the commutativity of the diagram.

It seems like it skips a step relating the two. It constructs a path in $\tilde X$ from $\tilde x_0$ to $\tilde x_1$ and then immediately starts the line $S = \dots$

What is the link between the constructed path, the conjugate groups, and the commutative diagram? (The figure at the bottom of the image is the diagram used in the proof)

enter image description here

Oliver G
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  • Well, I think that $\sigma(x)=[\lambda^{-1}]x[\lambda]$ is the canonical isomorphism and since $S=[\lambda^{-1}]H[\lambda]$ then $S=\sigma(H)$. That's where the first equality "$S=\ldots$" comes from. The second one comes from the commutative diagram (it would be easier if we knew what Theorem 10.10 says) since $\sigma p_{}=p_{}\Sigma$. – freakish Sep 07 '17 at 15:50

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I feel this a good example where one uses the algebraic model of a covering map in terms of a covering morphism $q: H \to G$ of groupoids.

Certainly if $p: \widetilde{X} \to X$ is a covering map of spaces then the induced morphism $\pi_1(p)$ of fundamental groupoids is a covering morphism. The definition of covering morphism $q: H \to G$ is that if $y \in Ob(H)$ and $g: p(y) \to x $ in $G$ then there is a unique $h: y \to z$ in $H$ such that $q(h) = g$.

The full details are in Chapter 10 of Topology and Groupoids, (T&G, downloadable), as they were in previous differently titled editions (1968, 1988). Peter May's "Concise ..." book also has some information on covering morphisms of groupoids. The simplification of notation, replacing $\pi_1(X,x)$ by the object group $G(x)$, should also make it easier to write down the necessary arguments.

The relation of covering morphisms to actions of groupoids on sets is given in this stackexchange answer, and also in T&G. The notion of covering morphism of groupoids is also a special case of the useful notion of fibration of groupoids.

Ronnie Brown
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