Clarifying's Rotman's proof of $\pi(K, p) \cong G_{K,T}$

Question

I understand most of this proof, but there's some definitions that are not completely clear to me. I'll provide only the relevant snippet of the proof the confuses me. Above the proof, I provide relevant definitions.

1. Definitions and Proof

2. Rotman's alternative definition of ker from an earlier chapter

3. Questions

1. What exactly is $F$? As a free group with basis all edges $e \in K$, this means $F$ is all reduced words of the form $e_1^{q_1} \cdots e_n^{q_n}$. But note that these need not be edge paths in $K$. If we required edge paths only, I don't believe F would be a free group. But then $\text{im} \phi \not\subset \pi(K,p)$, because you have these non-edge paths in the domain.

2. If you have a relation $(x = y)$, my understanding is this means $x y^{-1} \in R$. To show that type (ii) relations are in $\text ker \phi$, Rotman shows $\phi((u,v)(v,w)) = \phi((u,w))$. If you already assume that $\phi$ is a homomorphism, then that equality suffices. I think that's what he's doing, but please confirm. (Rotman is trying to show that $\phi$ induces another homomorphism, so his language seemed a bit imprecise.)

Answer 1

1. As Rotman says, $F$ is the free group on the edges of $K$. Because $F$ is free, to define a homomorphism from $F$ to another group, it suffices to say what it does on each generator, i.e. on each edge of $K$, and that's what he does. For example, $\varphi((u,v) (x,y))$ is by definition $\varphi((u,v)) \varphi((x,y))$, and this makes sense even if the two edges $(u,v)$ and $(x,y)$ do not share a vertex. Since $\varphi((u,v))$ is a loop based at $p$ for each generator, the same is true when you apply $\varphi$ to a product of edges in $F$: it's just the concatenation of the corresponding loops.

   Edit: the first part of the previous paragraph could be phrased better. For any group, if you define what a homomorphism does on the generators, then that determines the homomorphism overall. For a free group, furthermore, you can specify any targets you want for the generators and you get a valid homomorphism; for a non-free group, this may not be the case. For example, you can send $1 \in \mathbb{Z}$ wherever you want to define a group homomorphism $\mathbb{Z} \to G$, but you can't do this with $1 \in \mathbb{Z}/2\mathbb{Z}$ to define $\mathbb{Z}/2\mathbb{Z} \to G$: there are restrictions on where $1$ can go to get a valid group homomorphism.

2. $\varphi$ is defined on the generators of $F$, and since $F$ is free, this automatically defines a homomorphism.

Answer 2

Maybe a picture would be helpful for intuition? Consider the following two-dimensional complex below with a loop drawn. Notice how the red loop is homotopic to the concatenation of the blue paths. Since each simplex is simply connected we can homotope our paths to move along the edges. Imagine taking the red loop and inflating it until it encircles the edges. In the end we can represent each homotopy class by paths which are generated by edges in $K$.