proof of the injectivity of an application I have to solve the following exercise ?
Let $G$ and $H$ be two groups, $G = \langle g_1,\ldots,g_r\rangle$. I have the following map :
$$\begin{align*}
f \colon \mathrm{Hom}(G,H) &\longrightarrow H^r,\\
\phi &\longmapsto (\phi(g_1),\ldots,\phi(g_r)).
\end{align*}$$
I have to show that it is an injection; so I tried to find the kernel of the map but, my equation is :
$(\phi(g_1),\ldots,\phi(g_r)) = (e_H,\ldots,e_H)$.
But I don't see how i can conclude if $\phi$ is not an injection. I suppose it has a link with the definition of $G$ but I have a hard time understanding it's structure. An element $g \in  G$ can of course be $ g = (g_i)^n, i \in \{1,\ldots,r\}$ but can it be $g = g_i*g_j, j \neq i$ ?
Can you help me please ?
 A: Though the question was originally tagged "abelian groups", you seem to be saying in the comments that $G$ and $H$ are not assumed to be abelian groups. If that is indeed the case, then note that $\mathrm{Hom}(G,H)$ is not usually a group. It's not a group under composition, clearly, but it is also not a group under pointwise multiplication: if $\phi,\psi\colon G\to H$ are group homomorphisms, then in general $g(ab)h(ab) = g(a)g(b)h(a)h(b)$ is not equal to $g(a)h(a)g(b)h(b)$. (It is a group under pointwise multiplication when $H$ is abelian). Given that you cannot assume your set is a group, checking what elements map to the identity may tell you the map is not injective (if you can find two things that do), but it will not decide the issue: you can have a function that is not one-to-one but where only one thing goes to the identity.
The key here is just that if $G$ is generated by $X\subseteq G$, then two homomorphisms $\phi,\psi\colon G\to H$ are equal if and only if they agree on $X$.
One direction is clear: if $\phi=\psi$, then $\phi(x)=\psi(x)$ for all $x\in X$.
Conversely, if $\phi(x)=\psi(x)$ for all $x\in X$, then let $g\in G$. Since $G=\langle X\rangle$, there exist $x_1,\ldots,x_m\in X$ and $\epsilon_1,\ldots,\epsilon_m\in \{1,-1\}$ such that
$$ g = x_1^{\epsilon_1}\cdots x_m^{\epsilon_m}.$$
Therefore,
$$\begin{align*}
\phi(g) &= \phi(x_1^{\epsilon_1}\cdots x_m^{\epsilon_m})\\
&= \phi(x_1)^{\epsilon_1}\cdots \phi(x_m)^{\epsilon_m}\\
&= \psi(x_1)^{\epsilon_1}\cdots \psi(x_m)^{\epsilon_m}\\
&= \psi(x_1^{\epsilon_1}\cdots x_m^{\epsilon_m})\\
&= \psi(g).
\end{align*}$$
From this, you should be able to prove that if $f(\psi)=f(\phi)$, then $\psi=\phi$. Again, you can't just see whether $f(\psi)=(e,e,\ldots,e)$, unless you are dealing with the pointwise group structure of $\mathrm{Hom}(G,H)$, which you are only guaranteed is a group structure when $H$ is abelian. (Though for some pairs of groups it may work; e.g., if $G=S_5$ and $H=C_2\times K$, where $K$ is the Heisenberg group of order $3^3$; but in general, it won't.)
