To illustrate, let the generators be AB/A and ABAB. Let G be the figure 8, with a loop for A and a loop for B. Map a triangle onto AB/A, and a square onto ABAB. Join the triangle and square at a base point x. Call this graph G0, with the generators of π1(G0) mapping onto the generators of H. The map on the star of x is not injective, because three edges leave x and map to a. Fold these edges into 1, and merge the three terminal vertices. The result is a square with a B loop hanging off of one corner. Call this graph G1, and note that G0 maps onto G1, maps into G. At the same time, G0 maps into G, giving a commutative diagram.
Repeat this process. Each time we lose atleast one edge and vertex, so after finitely many iterations, we reach Gn, which is star injective into G. Continuing our example, two edges leave a vertex and map to B. One is the loop and one is the side of the square. Merge these together and find an AAB triangle with a B loop at the AA corner. This is star injective into G, and the fundamental group embeds. This group is the same as H, though that is less than obvious. Let's take it one step at a time.
going back to G0, a word is in H iff it starts and ends at x. In other words, every path back to start is a word in H. This is clear by construction.
Move from one graph to the next by merging two edges together. Let y be the center of the star that is not injective, and let u and v be the two endpoints. By induction, every path back to x is a word in H. If u = v then merging the edges together will not create any new paths back to x, so assume u ≠ v. These vertices become one, and this could create a new path back to x if there was formerly a path from x to u and a path from v to x. Use these paths as follows. Build a closed path from x to u, to y and back to x. Build another closed path from x to y to v, and back to x. These are both words in H. concatenate them, and find a word in H that represents the path from x to uv and back to x. this works even if u or v is equal to y. Do this for each new closed path created by the merger. For each Gi, the paths back to start are the words of H, and the same holds for Gn. Select a basis for Gn, and you have a basis for H. This is usually done by building a spanning tree, and assigning a generator to each chord.
The map from G0 onto Gn induces a homomorphism on the fundamental group. As shown above, this is an epimorphism. Both groups are free, hence the domain cannot have lesser rank. Therefore the rank of H is at most the number of generators.
Return to our example, where Gn is an AAB triangle with an extra B loop hanging off the AA corner. Choose AA as our spanning tree. Remember, words are written relative to x. Thus the cycles are AAB and AB/A. Verify that AAB is accessible from the original generators, and at the same time, AB/A * AAB gives ABAB back again.
To test whether a word is in H, lift the path up to Gn. If a lift is possible, and the lift returns to x, the word belongs to H. The lift is unique, and gives the representation of the word with respect to the free basis of H.
For another example, review the subgroup of index 2, and build the graph that supports these generators. After it is folded, you have three circles, with the base point joining the bottom circle and the middle circle. The bottom circle and the top circle are loops that map to B. The right side of the middle circle maps to A, and the left side of the middle circle maps to A (both going counterclockwise). Yes, the graph is star injective. Let A, traveling up the right side, be the spanning tree, and find the generators AA, B, and AB/A. H is free on these three generators, though it was also free on AA, B, and ABA.
Here is another example, which was referenced near the end of the theorem on the fundamental group of a 1 complex. Let H live in the free group on A B and C, with generators AB, BC, and BAC. G0 consists of three closed paths joined at a base point x. Clearly BAC needs to merge with BC, giving the closed path BC with a loop A at the point opposite to x. This graph is star injective into G, so we're done. If you lift the path A, you do not get back to x, hence A is not in H.
If two distinct edges A leave u1u2, headed to different vertices, then either G1 has two A edges leaving u1, or G2 has two A edges leaving u2. Since G1 and G2 are star injective into G, W is star injective into G.
A path, starting at x, returns to x, iff the corresponding paths in G1 and G2 start and end at their base points. The words of W are H1 intersect H2. These can be expressed using a basis, which is finite, since the vertices and edges of W are finite.
Let's try an example. Assume H1 is generated by AB and BA, and H2 is generated by ABA and B. Build the graphs G1 and G2, and note that they are already folded. Each graph has 3 vertices, so W has 9 vertices. Number the vertices of each graph as 1 2 and 3, and denote the vertices of W with 2 digits, as in 13, 22, etc. Now A maps 11 → 22, 13 → 21, 31 → 12, and 33 → 11. B maps 21 → 11, 22 → 13, 11 → 31, and 12 → 33. Note that 23 and 32 are isolated points. That leaves 7 points and 8 edges, which implies two cycles. Start at the root 11, and connecte to 21, 22, 31, and 33. Join 33 to 12, and 22 to 13. This is our spanning tree. The last two edges provide the basis: ABAB and BABA. This result is farily intuitive, but would be difficult to prove without these graphical techniques.