Trying to understand proof of isomorphsim between two group presentations I am trying to understand a proof given of an isomorphism between an infinite and finite presentation of Thompson's group F in the following paper by Cannon, Floyd and Parry. 
http://book-on-words.googlecode.com/svn/trunk/cannonfloydparryThompson.pdf
I'm referring to Theorem 3.1 on page 7.
So, I understand how to construct the surjective homomorphism (I'll just call it $\phi$) mentioned at the beginning. It's the steps after this that I don't quite understand. How does:
$(i)$ Showing that the defining relations of $F_1$ are contained in the kernel of $\phi$
$(ii)$ Showing that there exists a homomorphism the other way (from $F_2$ to $F_1$)
Prove the result?
To show $\phi$ is injective surely you have to show that the defining relations of $F_1$ are equal to ker($\phi$)? (since then $\phi$ has trivial kernel)
Also, since $\phi$ is defined by $\phi(A) = X_0$ and $\phi(B) = X_1$, does showing there exists a homomorphism the other way that maps $X_0$ to $A$ and $X_1$ to $B$ not prove that $\phi$ has an inverse, and so must be an isomorphism anyway? 
If so, why bother to show $(i)$?
Thanks in advance for any help
 A: Let me explain the general framework of the proof.
Let $G_1 = \langle X_1 \mid R_1 \rangle$ be a group presentation, where $R_1$ is a set of relators. So $G_1 = F_1/\langle R_1^{F_1} \rangle$, where $F_1$ is the free group on $X_1$.
Now if $H$ is any group, and $\phi:X_1 \to H$ is any map, then $\phi$ extends uniquely to a homomorphism $F_1 \to H$ and, if $R_1 \le \ker \phi$, then $\phi$ induces a homomorphism $\bar{\phi}_1:G_1 \to H$.
In this proof, we have another group presentation $G_2 = \langle X_2 \mid R_2 \rangle$, and the homomorphism $\bar{\phi}_1:G_1 \to G_2$ is defined as above.
Then the equivalent thing is done in the other direction, and a homomorphism $\bar{\phi}_2:G_2 \to G_1$ is defined.
It is straightforward to check in this example that $\bar{\phi}_2\bar{\phi}_1(x)=x$ for all $x \in X_1$ and $\bar{\phi}_1\bar{\phi}_2(x)=x$ for all $x \in X_2$. This proves that $\bar{\phi}_2\bar{\phi}_1:G_1 \to G_1$ and
$\bar{\phi}_1\bar{\phi}_2:G_2 \to G_2$ are both equal to the identity maps. So $\bar{\phi}_1$ and $\bar{\phi}_2$ are mutually inverse maps, and hence both are isomorphisms.
