If \(X\) is a finite graph and \(Y\) is a subgraph homeomorphic to \(S^1\) and containing the basepoint \(x_0\), show that \(\pi_1(X, x_0)\) has a basis in which one element is represented by the loop \(Y\).
Solution
Let \(Y\subseteq X\) be the given subgraph homeomorphic to \(S^1\) and containing the basepoint \(x_0\). Choose an edge \[ e\subseteq Y. \]
1. The subgraph\[ Y\setminus \mathring e \]is a tree.
Removing the interior of one edge from a circle leaves a closed interval subdivided into finitely many edges. Thus it is connected and contains no cycle.
2. Extend\[ Y\setminus \mathring e \]to a maximal tree\[ T\subseteq X. \]
Since \(X\) is finite, repeatedly adjoin edges that connect the current tree to a new vertex or otherwise enlarge it without creating a cycle. The process terminates at a maximal tree containing the original subgraph.
3. The edge \(e\) does not lie in \(T\).
If \(e\) also belonged to \(T\), then \(T\) would contain all of \(Y\). But \(Y\) is a cycle, contradicting that \(T\) is a tree.
4. The standard free basis of \(\pi_1(X,x_0)\) associated to the maximal tree \(T\) contains the loop \(Y\).
For every edge \(f\) of \(X\setminus T\), the associated basis loop starts at \(x_0\), follows the unique path in \(T\) to one endpoint of \(f\), traverses \(f\), then returns to \(x_0\) along the unique tree path from the other endpoint.
Take \(f=e\). Because \[ Y\setminus\mathring e\subseteq T, \] the unique path in \(T\) between the endpoints of \(e\) is exactly the complementary arc of \(Y\). Hence the basis loop associated to \(e\) traverses precisely the circle \(Y\), up to orientation. Thus one basis element is represented by \(Y\).