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4 fixed grammar

$EDIT$ I get a mistake simply gived a answer to another question (sorry, mistake), I find the answer as exercises in "Rings os Quotients" B. Strenstrom, Springer Verlag 1975, pag 134 n.13.

I try to do a proof (I hope):

THEOREM) for a abject $C$ in a Grothendieck category (we think simply to a module category) the following are equivalent:

1) For any sequence of subobject like $C_1\subset C_2\subset\ldots C$ we have $C=C_m$ for some $m$.

2) For any sequence of subobject like $M_1\subset M_2\subset\ldots M$ with union $M$ we have that $(C, M)=\cup_n(C, M_n)$ (naturally).

3) The functor $(C, -)$ commute by denumerable coproducts.

4) The functor $(C, -)$ commute by coproducts.

5) The functor $(C, -)$ commute by directed unions (i.e. $C$ is $f.g.$).

PROOF. $(1\Rightarrow 2):$ we have to proof that any $f: C\to M$ has image in some $M_m$, if we put $C_n:=f^{-1}(M_n)$ we done.

$(2\Rightarrow 3):$ of course $(C,-)$ commute by finite coproduts (they are biproducts), we have to prove that a $f: C\to M$, with $M=\coprod_n X_n$ as a factorization on a finite summands, let $M_n:=\coprod_{i\leq n} X_i$ we done.

$(3\Rightarrow 4):$ we have to prove that a $f: C\to M$, with $M=\coprod_{i\in I} X_i$ as a factorization on a finite summands, suppose the opposite: then we have an infinite denumerable set of indices $i_0, i_1\ldots \in I$ such that for any integer $n$ exist a $x_n\in C$ with $f(x_n)_{i_n}\neq 0$

then we consider $J:= I\setminus${$i_0, i_1\ldots$} and the quotient map $\pi: \coprod_{i\in I}X_i\to (\coprod_{i\in I}X_i)/(\coprod_{j\in J}X_j)\cong\coprod_n X_{i_n}$ and the composition $\pi\circ f: C\to \coprod_n X_{i_n}$, this map isnt factorizable to a finite summands (absurd).

$(4\Rightarrow 5):$ Let $M=\cup_{i\in I} M_i$ where $I$ is a directed order. We can suppose $I$ cofinite i.e. for any $i\in I$ exist only finite $j$ such that $j\leq i$ (e.g. "Shape Theory" Sibe MArdiesic NH 1982 T.2 pag. 10). Then the natural map $\pi: M\to \coprod_{i\in I} M/M_i$ with $(\pi(x))_i=\pi_i(x)$, $\pi_i: M\to M/M_i$ natural, is well defined. We have to proof that any $f: C\to M$ has a image on some $M_j$, considering $\pi\circ f: C\to \coprod_i M/M_i$ then this map has a factorization on finite summands $M/M_{i_1},\ldots M/M_{i_N}$, if some $M_{i_n}$ is $M$ the assert is trivial

if no we can have a $j\in I$ strictly greater of any $i_1,\ldots i_N$, then $f\circ \pi_j: C\to M\to M/M_j$ is the $0$ map, then the image of $f$ is in $M_j$.

$(5\Rightarrow 2):$ Trivial

$(2\Rightarrow 1):$ Let $M:=M,\ M_n:=C_n$ and considerind $1_C$

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From "Abelian CAtegories and its application to Rings and Modules " by Popescu N, par. 3.5 pag 88:

Popescu using the names "small" and "finitely presented" as for the yours "sumcompact" and "compact" respectively.

Popescu call a object $X$ (of a Grothendick abelian category $\mathcal{C}$)
of "finite type" is if for any direct union of subobjets $Y=\cup_{i\in I}Y_i$ the natural morphism $Colim_{i\in I} \mathcal{C}(X, Y_i)\to \mathcal{C}(X, Y))$ is a isomorphism, this is equivalent to:

for any directed union of subobjets $X=\cup_{i\in I}X_i$ there is a $i_o\in I$ such that $X=X_{i_0}$.

In a category of modules finitely presented is equivalent to the usal definition (there is a exat exact $0\to A\to X\to C\to 0$ with $A,\ B$ finitely generated), and finite type is equivalent to finitely generated.

From 5.4 of Popescu book a finitely generated objact module is small (sumcompact). And of course exist finitely generated modules that aren't finitely presented. then we have the implications:

finitely presented $\Rightarrow$ finitely generated $\Rightarrow$ small (suncompact) and finitely .generated$\not\Rightarrow$ finitely presented

Then cannot have that small (sumcompact)$\Rightarrow$ finitely presented.

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