Is it possible to check injectivity of a homomorphism by checking just the generators of the group?

Question

Let $G$ be a group with generators $\{g_i\}_{i\in I}$.

Let $\psi:G\to H$ be a group homomorphism (to another group $H$).

Is it possible to check injectivity of $\psi$ just by checking the generators?

That is, if $\psi(g_i)=\psi(g_j)$ implies $g_i=g_j$ for all $i$, $j$, does it necessarily mean that $\psi$ is injective?

Thanks.

Answer 1

No. Take $\varphi\colon\mathbb{Z}^2\longrightarrow\mathbb Z$ define by $\varphi(x,y)=x+2y$. The group $\mathbb{Z}^2$ is generated by $(1,0)$ and $(0,1)$ and $\varphi(1,0)\neq\varphi(0,1)$. However, $\varphi$ is not injective.

Answer 2

No. With what you suggest, the map $\Psi: G\to G$ that is the identity on all generators except one, $g_1$ which is mapped to $e$, would be injective.

Or if for instance every $g_i$ is of infinite order you could define $\phi(g_i)=g_1^i$, which is again strongly not injective but fails your test.