What does it mean for the fundamental group of the n-leafed rose $C_{n}$ to be free of rank $n$?

Question

The derivation from "introduction to knot theory" Richard H. Crowell is given below:

My questions are:

1-What does it mean for the fundamental group of the n-leafed rose $C_{n}$ to be free of rank $n$?

2-Why if a group is infinite cyclic, it is called free group of rank 1? 3- Why we consider The open neighborhood of p to consist of $2(n + 1)$ disjoint, open arcs, why specifically $2(n+1)$, is it because each leaf can be divided into 2 arcs?

4-why in figure 23, part of it is drawn dotted?

Could anyone help me in answering these questions please?

EDIT:

Answer 1

First I'll say that an essential prerequisite to this question is knowledge of free groups. Granted, many people's first experience with free groups arises when they first encounter Van Kampen's Theorem. Nonetheless, what one should do at that point is to set aside Van Kampen's Theorem and learn about free groups. The essential features are this:

• Given a group $G$, a subset $S \subset G$ is a free basis of $G$ if for any group $H$ and any function $f : S \to H$, there exists a unique homomorphism $F : G \to H$ such that $f = F \mid S$.
• A group is free if it has a free basis.
• If a group is free than all of its free bases have the same cardinality, and that number is called the rank of the free group.

There are further very important features regarding uniqueness of a reduced word in the elements of $S$ and their inverses for each element of $G$, but that is not essential for this problem.

To answer your question 1, the statement of the theorem being proved in this passage of Crowell is that $\pi_1(C_{(n)})$ is a free group of rank $n$ and $\{\omega_1 x_1,...,\omega_n x_n\}$ is a free basis.

To answer your question 2, in the group $\mathbb Z$ the set $\{1\}$ is a free basis, as one can easily prove. Therefore $\mathbb Z$ is a rank $1$ free group. Every infinite cyclic group is therefore a rank $1$ free group: a group is infinite cyclic if and only if it is isomorphic to $\mathbb Z$; and the property that a group is "free of rank $n$" is an isomorphism invariant.

To answer your questions 3 and 4, as one can see from reading the statement of Van Kampen's Theorem, open subsets play an important role. The difference between dotted and undotted lines is meant to emphasize issues explained in the text that you pasted in, regarding various subsets of $C_{(n+1)}$ that are and are not open; this is needed for purposes of applying Van Kampen's Theorem. The solid lines are meant to represent an open subset containing $p$ which is denoted $N$; you can see this from the diagram, by observing the pointers from the letter $N$ to various portions of that open subset. The dotted lines are meant to represent the complement of $N$.

Answer 2

1) It is the free group on $n$ generators $F_n = \langle g_1,\dots,g_n \mid \; \rangle$, where each generator $g_i$ is given by the homotopy class of the loop going around $X_i$. I can give you concrete representatives of the isomorphism class of $F_n$ for some given $n$, namely for example the integers $\mathbb{Z}$ for $n =1$ or $\langle a,b \rangle \subset \text{GL}_2(\mathbb{C})$, where $$a = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}, \;\; b = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix},$$ for $n =2$. Otherwise it is just some abstract group satisfying some universal property or you can construct some representative in the way that Rylee said in the comment.

One usually proves that fact by using the Seifert van Kampen theorem which is probably related to some of the statements the author is quoting.

2) See 1).

3) Yes exactly. Each of the circles has two arcs starting at $p$.

4) I guess to visualize that we excluded the dotted part when we took the neighborhood of $p$.

Answer 3

There is another way to consider this, cf Topology and Groupoids (T&G), or Categories and Groupoids. Define the fundamental groupoid $G= \pi_1(X,S)$ on a set $S$ of base points. For example, let $X$ be the interval $[0,n]$ and let $S$ be the set $\{0,1,\ldots, n\}$. Since $[0,n]$ is convex, the groupoid $G$ has then $n+1$ objects, and exactly one arrow say $\tau_{ij}:j \to i $. This is a rather easy kind of free groupoid, called a tree groupoid in T&G. Note that $\tau_{ij}\tau_{jk}=\tau_{ik}$.

$$ i \xrightarrow{\tau_{ij}}j \xrightarrow{\tau_{jk}} k $$

Now if you identify all the points of $S$ to a single point $p$, then the space $X$ becomes the $n$-leaf rose. and its fundamental group at the point $p$ is the free group on the images of the $\tau_{i,i+1}: i \to i+1$ for $i=0, \ldots, n-1$. You then need a version of the van Kampen Theorem for the fundamental groupoid on a set of base points to get the expected result, and of course many more.

Compare this mathoverflow discussion.

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