I was trying to define a homomorphism between some finite groups and I had the following idea.
Suppose that $G = \langle a, b\rangle $, and $H = \langle x, y\rangle$. We define $\varphi:G \to H$ by $\varphi(a)=x,$ $\varphi(b)=y$, and for an arbitrary element of $G$, $g_1^{\epsilon_1}g_2^{\epsilon_2} \cdots g_n^{\epsilon_n}$ (where each $g_i$ is either $a$ or $b$, and each $\epsilon_i$ is either $1$ or $-1$), then $\varphi(g_1^{\epsilon_1}g_2^{\epsilon_2} \cdots g_n^{\epsilon_n}) = \varphi(g_1)^{\epsilon_1} \varphi(g_2)^{\epsilon_2} \cdots \varphi(g_n)^{\epsilon_n}$.
If this is well-defined, then this is obviously a homomorphism. However, I do not know if this is well defined. I suspect not.
Best Answer
Any arbitrary map of generators to generators need not extend to a homomorphism, even if there is only one generator. For example, there is no nontrivial homomorphism between two cyclic groups of different prime orders.
The problem is that any combination of generators that results in the identity in the original group must also evaluate to the identity when you map to the target generators. Otherwise the map would not be a homomorphism, or even well defined (so actually not a function at all).