Need help understanding a step in a proof in Grothendieck's Tohoku So I have been trying to work through Grothendieck's Tohoku paper. A copy of the English translation can be found here. I am just stuck on one step in the proof of Proposition 1.9.1 (page 14 of the paper or page 23 of the PDF). In particular, I am not comfortable with (possibly set theoretic issues with) how he shows equivalence of (2) and (3).
The setup is as follows: For a category $\mathcal{C}$, let $(U_{i})$ be a family of objects of $\mathcal{C}$. We say that the family $(U_{i})$ is a $\textit{family of generators}$ if for any object $A$ of $\mathcal{C}$ and any subobject $B \hookrightarrow A$ (via some monomorphism $m$) and any index $i$, any morphism $U_{i} \longrightarrow A$ factors via $m$.
Now let $\mathcal{C}$ be an abelian category admitting arbitrary coproducts (direct sums in Grothendieck's terminology). Let $(U_{i})$ be a family of objects of $\mathcal{C}$ and defined the coproduct
$$
U = \bigoplus_{i} U_{i}.
$$
The claim is that the following conditions are equivalent:
1) $(U_{i})_{i \in I}$ is a family of generators of $\mathcal{C}$.
2) $U$ is a generator of $\mathcal{C}$.
3) Any object $A$ of $\mathcal{C}$ is isomorphic to a quotient of a direct sum $U^{\oplus J}$ of objects that are all isomorphic to $U$.
My main issue is understanding how my intuition of "image" of a morphism (in terms of elements of the target object that get mapped to) aligns with the definition in terms of the kernel of cokernel. Thus when Grothendieck constructs a morphism 
$$ U^{\oplus I} \longrightarrow A $$
where the index set $I$ is the hom set $\text{Hom}(U, A)$, he is able to instantly conclude that the image of the morphism must be $A$ itself which I struggle to see. 
Would someone be able to expand a bit on the details of how he (or if there is another proof that would be welcomed to) gets equivalence between (b) and (c)?
Any help is appreciated
Thanks
 A: Your definition of generating set isn't properly cited. In the original text it is:

We say that it is a family
  of generators of $C$ if for any object $A\in C$ and any subobject $B \ne A$, we can find an $i\in I$
  and a morphism $u : U_i\to A$ which does not come from a morphism $U_i\to
B$.

Here $B\ne A$ wants to mean that $B$ is a proper subobject, i.e. the corresponding monomorphism is not isomorphism. Besides, '$u$ does not come from a morphism $U_i\to B\,$' means, using the monomorphism $m:B\hookrightarrow A$, that for any $v:U_i\to B$ we have $u\ne m\circ v$.
The image of a morphism $f:U\to A$ can be defined as the smallest subobject of $A$ through which $f$ factors.
Let $B$ denote the image of $\phi:U^{\oplus I}\to A$. By $U$ being generator, if $B$ is a proper subobject, there is an $u:U\to A$ which does not factor through $B\hookrightarrow A$. 
Now, we happened to have $I=\hom(U,A)$ here, so $u\in I$ and - as Roland commented - we can write $u=\phi\circ\iota_u\ $ where $\iota_u:U\hookrightarrow U^{\oplus I}$ is the inclusion at index $u$. By the above property of image, $\phi$ factors through $B\hookrightarrow A$, thus, so does $u$. $\,$ Contradiction.
