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This is about exercise V.4.1. in Massey's A Basic Course in Algebraic Topology. Section V.4. is about the fundamental group of a covering space. Massey assumes that all spaces involved are both arc-connected and locally arc-connected.

Theorem 4.1. states that for a covering space $p: \tilde{X} \to X$, $\tilde{x}_0 \in \tilde{X}$ and $x_0 = p(\tilde{x}_0)$ the induced morphism $p_*:\pi(\tilde{X},\tilde{x}_0) \to \pi(X,x_0)$ is a monomorphism. What follows is a discussion on what happens if one replaces $\tilde{x}_0$ by another $\tilde{x}_1$ which is also mapped to $x_0$ via $p$. The result is that $p_*(\pi(\tilde{X},\tilde{x}_0))$ and $p_*(\pi(\tilde{X},\tilde{x}_1))$ are conjugate subgroups of $\pi(X,x_0)$ and that all subgroups of $\pi(X,x_0)$ which lie in the conjugacy class of $p_*(\pi(\tilde{X},\tilde{x}_0))$ arise this way. This is the content of

Theorem 4.2. Let $p:\tilde{X} \to X$ be a covering space and $x_0 \in X$. Then the subgroups $p_*(\pi(\tilde{X},\tilde{x}))$ for $\tilde{x} \in p^{-1}(x)$ are exactly a conjugacy class of subgroups of $\pi(X,x)$.

I'm having problems with

Exercise 4.1 Discuss the effect of changing the "base point" $x_0$ in the statement of Theorem 4.2. to a new base point $x_1 \in X$.

My first idea would be to pick some path connecting $x_0$ and $x_1$ in $X$ (it exists by the assumption of path-connectedness). This would yield an isomorphism of groups $\pi(X,x_0) \to \pi(X,x_1)$. Under this isomorphism the conjugacy class coming from Theorem 4.2. for $x_0$ maps to a conjugacy class of subgroups in $\pi(X,x_1)$. Is the content of the exercise to compare this class to the class of subgroups of $\pi(X,x_1)$ that I get when I use Theorem 4.2. with $x_1$ in place of $x_0$? If so, could anybody provide a hint please?

edit: Here is the diagram I have been talking about in the comments:

$\pi(\tilde{X},\tilde{x}_0) \longrightarrow \pi(\tilde{X},\tilde{x}_1)$

$\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$

$\pi(X,x_0) \longrightarrow\,\, \pi(X,x_1)$

The vertical maps are $p_*$, the lower horizontal one is conjucation with the class of a fixed path from $x_0$ to $x_1$ in $X$ and the upper horizontal map is conjugation with the class of the lift of said path.

Is there something else to this exercise? If I now change $\tilde{x}_0$ in the fiber $p^{-1}(x_0)$ and I then change the lifting of the path from $x_0$ to $x_1$ to $\tilde{X}$ accordingly (i.e. changing its starting point), then I run through a list of commutative squares as above, one for each $\tilde{x_0} \in p^{-1}(x_0)$. Doing so I run through the whole conjugacy class of $p_*\pi(\tilde{X},\tilde{x_0})$ in $\pi(X,x_0)$ in the left half of the diagram. Is the question now what happens on the right half of the diagram when I do so? My guess is that I simply run through the whole conjugacy class of $p_*\pi(\tilde{X},\tilde{x_1})$ in $\pi(X,x_1)$. Is that correct?

Did I make myself clear? I was fixing $\tilde{x_0}$ first and then left it run again, hopefully I did not lose you with that.

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From what you wrote I'd assume that you are, too, supposed to change the base point in $\tilde{X}$ to some point which is mapped to $x_1$. This can be acchieved by lifting your path from $x_0$ to $x_1$. –  user20266 Jun 5 '12 at 16:38
    
Apparently I cannot post an answer to my own question. I understand what you mean: I start with a homotopy class of paths from $x_0$ to $x_1$ in $X$, then by a lemma of section 3 of that chapter I can lift this class to a unique class of paths in $\tilde{X}$ starting at the fixed $\tilde{x}_0 \in p^{-1}(x_0)$. Then I let $\tilde{x}_1$ be the endpoint of that lifted class of paths which is an element of $p^{-1}(x_1)$. Then I get a commutative square of groups involving $p_*$ as the vertical maps, the conjugation by the lifted class as the upper horizontal map and the conjugation ... –  Lennart Jun 6 '12 at 14:29
    
... by the original class as the lower horizontal map. Is that all there is to it? Or am I asked to do more (I cannot draw the diagram in the comments and I cannot post an answer, so I will edit my question in a second). –  Lennart Jun 6 '12 at 14:30
    
As a side remark: you should be able to answer your own question. There should be a button at the bottom of the screen. –  user20266 Jun 6 '12 at 17:11
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2 Answers

up vote 1 down vote accepted

Following up my comment which is now below (!), here is a solution in terms of groupoids.

Let $p: H \to G$ be a covering morphism of groupoids, let $x' \in Ob(H)$ and let $p(x')=x$. Let $k \in G(x,y)$. Then $k$ has a unique lift $l \in H(x',y')$, say, (i.e. $p(l)= k$) since $p$ is a covering morphism, and of course $p(y')=y$. If $h \in H(x')$, and $p(h)=g$, then

$$ p(l^{-1}h l)= k^{-1}gk $$ and this proves the required conjugacy.

Note: We write $H(x')$ for $H(x',x')$ and call it the vertex group, or object group, of $H$ at $x'$. I like the idea of using the simplest notation, e.g. replacing $\pi_1(\tilde{X}, \tilde{x})$ by $H$. This can help to find the underlying processes, and indeed part of the history of mathematics is the history of improved notation.

Of course the proof in the topological framework and the algebraic framework are essentially the same, but algebraic topology is about algebraic models.

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Thank you for your answer and sorry for commenting on it so late. I am glad you mentioned the fundamental groupoid here since I know a few basic things about category theory already and replacing the fundamental group by the fundamental groupoid has clarified things a lot for me. I will try to find a copy of your book, for the moment I am working with the book Grundkurs Topologie by Laures and Szymik, they also formulate things using the fundamental groupoid. –  Lennart Jul 3 '12 at 11:16
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Since you do not ask for a solution to the exercise but rather what the exercise may be about, I think this is an admissible answer (and writing an answer is easier than commenting due to the $\LaTeX$ preview....):

Actually, yes, I'd say there is something more to the exercise. If $\gamma$ is a curve joining $x_0$ to $x_1$ and $\sigma$ another one, then $\sigma^{-1}\circ \gamma$ defines an element $\alpha$ in $\pi(X, x_0)$, which, to make this an interesting question, you should assume to be nontrivial. Then the question arises what the relation of the induced isomorphisms at the bottom of your diagram is, depending on whether you are choosing $\gamma$ or $\sigma$ to define it.

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These questions and answers show the extra difficulty of not using the appropriate algebraic model of a covering map of spaces, namely a covering morphism of groupoids. This is a morphism $p: H \to G$ of groupoids such that for each $x' \in Ob(H)$ and each $g \in G$ starting at $px'$ there is a unique $h \in H$ starting at $x'$ such that $p(h)=g$. For more details, see my book Topology and groupoids, Chapter 10, see my web page. (This account was first published in 1968.) –  Ronnie Brown Jun 7 '12 at 9:57
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