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So I was reading some Geometric group theory and came across Stalling's folding of graphs. Now I am trying to use the folding idea to prove that every finitely generated subgroup of a free group is free. Now, I already know that when I fold, I get surjective maps between the fundamental groups.

To finish the proof, I just need to show that if a graph map $A \rightarrow B$ is an immersion, then the induced homomorphism on fundamental groups is injective. Now, if I think about it, it's enough to show that immersions send tight loops to tight loops. But I can't really find out a way to show that. What I am thinking is if some tight loops doesn't get sent to a tight loop then I need to show it's not a immersion, i.e there are edges with same initial vertex in $A$ that has same image in $B$. Can someone help with this?

Edit: As asked in the comment, I am editing to include the definitions. A tight loop based on any vertex v is a closed edge path that starts and ends at v and all the consecutive edges are different, i.e. it doesn't backtrack. An immersion is a map that cannot be factored into folds, so it is a map between A and B so that for any two edges having same initial vertex in A, their image is different in B.

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  • $\begingroup$ Please include in your post the definitions of "immersion" and of "tight loop"; assuming that your definitions are what I expect, then a simple direct proof is probably best. $\endgroup$
    – Lee Mosher
    Commented Jun 7 at 16:10
  • $\begingroup$ @LeeMosher yeah I have included the definitions. $\endgroup$
    – Rinkiny Ghatak
    Commented Jun 8 at 9:56

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Consider a tight loop in $A$, denoted as an edge path $e_1 \cdots e_K$ without backtracking, based at a vertex $v_0 \in A$. For $1 \le i < K$ the subpath $e_{i}e_{i+1}$ does not backtrack in $A$. At the vertex $v_i$ which is the common terminal vertex of $e_i$ and initial vertex of $e_{i+1}$, it follows that the terminal direction of $e_i$ and the initial direction of $e_{i+1}$ are different directions.

Now consider an immersion, denoted $f : A \mapsto B$. It follows that $f(e_i)$ is a path without backtracking. It also follows that $f(v_i)$ is the common terminal endpoint of $f(e_i)$ and initial endpoint of $f(e_{i+1})$. Finally, at the point $f(v_i)$: the terminal direction of $f(e_i)$ equals the $f$-image of the terminal direction of $e_i$; and the initial directdion of $f(e_{i+1})$ equals the $f$ image of the initial direction of $e_i$; and so those directions are different because $f$ is an immersion and because the terminal direction of $e_i$ is different from the initial direction of $e_{i+1}$.

This proves that the concatenation $f(e_1) .... f(e_K)$ is a path without backtracking, i.e. a tight loop.

This can also all be said much more succinctly: an immersion is the same thing as a local embedding $A \mapsto B$; a tight edge path is the same thing as a local embedding $[0,1] \mapsto A$ (with a restriction on the endpoints); and the composition $[0,1] \mapsto A \mapsto B$ of two local embeddings is a local embedding.

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  • $\begingroup$ I am little confused with the wording. What do you mean by initial direction? And final direction? $\endgroup$
    – Rinkiny Ghatak
    Commented Jun 8 at 15:52
  • $\begingroup$ Associated to each point $p$ of a graph $G$ is the set of directions of $G$ at $p$. The number of directions at $p$ is equal to the valence of $p$. Formally the set of directions is the set of germs of oriented paths with initial vertex $p$. If $p$ is a vertex, each direction at $p$ is represented by a unique oriented edge with initial vertex $p$. For terminal directions, just reverse the orientation of the path or edge. $\endgroup$
    – Lee Mosher
    Commented Jun 8 at 16:39

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