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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

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  • $\begingroup$ I should also emphasize that we are not assuming this is a concrete category in any way, so that we can't speak of "elements" inside the objects. This is really the thing that's adding the extra level of abstraction that is making it hard for me. $\endgroup$ – Joe Sep 10 '17 at 15:43
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    $\begingroup$ Your definition of family of generator seems very weird. This is probably the source of your confusion. If $B$ denote the image of $U^{\operatorname{Hom}(U,A)}\rightarrow A$, then any morphism $U\rightarrow A$ factors through $B$ which is (should be) contradictory with the definition of $U$ being a generator. $\endgroup$ – Roland Sep 10 '17 at 15:49
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    $\begingroup$ I should probably add that the morphism $\phi:U^{\oplus\operatorname{Hom}(U,A)}\rightarrow A$ is not a random one. This is the following : an element $x\in U^{\oplus\operatorname{Hom}(U,A)}$ is a family $(x_f)_f$ where $f$ runs in $\operatorname{Hom}(U,A)$ and all but a finite number of the $x_f$ are zero. Then $\phi((x_f)_f)= \sum_{f} f(x_f)$. So if $f:U\rightarrow A$ is any morphism, $f$ is also the composite $U\rightarrow U^{\oplus\operatorname{Hom}(U,A)}\rightarrow A$ where the first map is the inclusion into the summand corresponding to $f$. This is why $f$ factor through $im(\phi)$. $\endgroup$ – Roland Sep 10 '17 at 18:38
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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.

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