A few days ago, I reconstructed the proof that $BS(2,3)$ is a non-Hopfian group, but I looked at it today and am now convinced that $BS(2,3) \to BS(2,3)$ given by $<a,b> \to <a^2,b>$ is in fact an isomorphism. (Here, the generators of $BS(2,3)$ are $a$ and $b$ and the relation of $BS(2,3)$ is $b^{-1}a^2ba^{-3} = 1$.) Would someone please comment on my proof so I can rejoin the ranks of civilized mathematics again?
The map splits into two maps: The first is an inclusion-like mapping $I : BS(2,3) \to BS(4,6)$ given by $a \to a^2,b \to b$ and transforms the relation from $b^{-1}a^2ba^{-3}$ to $b^{-1}a^4ba^{-6}$. The second is a projection map $P: BS(4,6) \to BS(3,2)$ sending $a$ to $a$ and $b$ to $b$ but using the stronger relation of $b^{-1}a^2ba^{-3}$. So the map that we are proving is not an isomorphism is $I \circ A$ (in algebraic notation).
Indeed, the third isomorphism theorem (with the free group <a,b> in front) shows that $P$ is a quotient map with kernel $\left<b^{-1}a^2ba^{-3}\right>$ of size 2. The tricky part, of course, is noticing that $I \circ P(b^{-1}aba^{-1}) = b^{-1}a^2ba^{-2} = a$. But once we know that, we have all of the group generated by $a$ and $b$ are in the target of $I \circ P$, which is exactly $BS(2,3)$, so $I \circ P$ is surjective.
Sadly, there is no element $w \in BS(2,3)$ such that $I(w) = b^{-1}a^2ba^{-3}$, because both the target of $I$ and the relation of $BS(4,6)$ only use even powers of $a$. This means we cannot lift the kernel of $P$ to the kernel of $I \circ P$.
Clearly (in the mathematical sense of "most likely to be wrong"), $I$ is injective because it introduces no new relations. And the last paragraph shows that $P | Im(I)$ is injective, so we conclude that $I \circ P$ is injective. We have already shown $I \circ P$ is surjective, so $I \circ P$ is an isomorphism.
Comments, please?
Yes, it really is!
I still need to think harder about what the kernel of P really is, but in any event it contains the element k = $a^{-2}b^{-1}a^2ba^{-2}b^{-1}a^2ba^{-2}$, because the relation for $BS(2,3)$ can be written as $b^{-1}a^2b = a^3$, and two simple substitutions show $k = a^{-2}a^3a^{-2}a^3a^{-2} = 1$. It's obviously in the target of $I$, because it is directly $k = I(a^{-1}b^{-1}aba^{-1}b^{-1}aba^{-1})$ -- I found this cool word in two Stack Overflow answers with examples of non-Hopfian groups which I cannot now locate, so I cannot cite them, but they are the answers with the correct classical proof.
Finally, $k \neq 1$ in $BS(4,6)$: If you look at the Cayley graph of $BS(4,6)$ in a direction where the $a$ arrows vanish, certain infinite sets of $b$ arrows combine into single arrows, and these resulting arrows form a tree of sorts: a (weak) simple regular directed graph where each vertex has 4 arrows pointing into it and 6 arrows pointing into it. By walking the Cayley graph using the path associated with the word that we use to represent $k$, it is clear that each $b$ and $b^{-1}$ moves in different directions and never double back, showing that $k$ has a b-word-metric distance of exactly 4, so $k$ cannot be a member of $\left<a\right>$, let alone 1.
Okay, this is too informal and the proof will have to wait for a later time, but it works for me. One of the other proofs on Stack Exchange cite "properties of HNN extensions" to show $k$ is non-trivial, so I don't feel too bad about using a goo of math terms to claim $k$ is non-trivial.
Anyhow, this gives us the answer that BS(2,3) is non-Hopfian, agreeing with the literature, and so I am back in a happy place today.