A pleasant result in additive categories

[killing_buddha]

Se $\mathcal C$ e' una categoria con tutti i limiti e colimiti finiti, dove ogni quadrato commutativo che sia un pullback e' anche un pushout e viceversa (indichiamo questi quadrati universali col nome di pullout), allora in essa

1. Esiste un oggetto zero, ovvero esiste un morfismo \(1\to \varnothing\) tra l'oggetto terminale e l'iniziale;
2. Esistono i biprodotti, ovvero \(X\times Y\cong X\amalg Y\) binaturalmente in $X,Y$.

Sono volutamente sloppy nell'enunciare il risultato, perche' confido che non esista solo un modo di dimostrare questa cosa (vera). La cosa piu' inaspettata e' che da questo si dovrebbe poter dedurre che una siffatta categoria deve essere additiva (ossia ogni $\hom(X,Y)$ supporta una struttura di gruppo abeliano).

C'e' un inganno, voluto, nel testo, che palesero' a tempo debito (e che non dovrebbe inficiare la dimostrazione): nel caso, scoprirlo e' parte della domanda.

[ficus2002]

1) Let $f,g:A\to B$ be two morphisms and let $e:E\to A$ be an equalizer of $f$ and $g$. We claim that there exists an isomorphism $h:E\to O$ ($O$ denote an initial object) such that $e=h\alpha_A$ where $\alpha_A:O\to A$ is the unique such morphism.

Let $e^\prime:B\to E^\prime$ the coequalizer of $f$ and $g$. The following square is a pushout, hence also a pullback:
$$
\begin{matrix}
A\cup A & \to & B\\
\downarrow & & \downarrow\\
A & \to & E^\prime\\
\end{matrix}
$$
where the upper arrow is induced by $f$ and $g$, the arrow on the left is induced by identity $1_A:A\to A$ and the right-handed arrow is $e^\prime$.
Let $e_i=e\text{ in}_i:E\to A\cup A$ where $\text{in}_i:A\to A\cup A$ denote the inclusion. Then both $e_1$ and $e_2$ serves as comparison morphisms for the square above. The unicity of pullback property implies $e_1=e_2$.
Since the following square is also a pusout, hence a pullback, if follows that $e$ factor througout $\alpha_A:O\to A$. $$
\begin{matrix}
O & \to & A\\
\downarrow & & \downarrow\\
A & \to & A\cup A\\
\end{matrix}
$$
Thus there exists $h:E\to O$ such that $e=h\alpha_A$. That $h$ is split epic because $O$ is initial and $e$, hence $h$, is monic beacuse $e$ is an equalizer. Consequently, $h$ is an isomorphism.

Let $I$ be a terminal object. Since $1_I:I\to I$ is the equalizer of $1_I$ and $1_I$ it follows that $1_I=h\alpha_I$ and $h:I\to O$.
2) For any object, since $1_A:A\to A$ is the equalizer of $1_A$ and $1_A$ we obtain an isomorphism $h:A\to O$.
In other word each object is a zero object. In particular, for any $X$, $Y$, the objects $X\times Y$ and $X\cup Y$ are isomorphic.

[killing_buddha]

Un metodo un po' circonvoluto, ma mi sembra corretto (ed e' anche un piacere veder rispondere te, ricordavo avessi interessi piuttosto diversi!).

L'inganno e' che una categoria con questa proprieta' deve per forza essere banale (e' equivalente alla categoria terminale): questo potrebbe essere un rilancio.
La nozione diventa importante (direi essenziale) qualora uno invece consideri strutture un attimo piu' generali, dove non e' piu' vero (e questo e' un secondo rilancio).

[ficus2002]

Se possiedi una dimostrazione più elegante la leggo volentieri.

[killing_buddha]

Si', in effetti la possiedo:

Let \(\varnothing\to 1\) be the canonical arrow between the initial and terminal object of \(\mathbf C\), which exists for sure since \(\mathbf C\) has finite limits.

We can build the pullout squares

[tex]\xymatrix{
\varnothing \ar[r]\ar[d] & 1\ar[d] & F\ar[r]\ar[d] & \varnothing \ar[d]\\
1 \ar[r] & C & \varnothing \ar[r] & 1
}[/tex][/list:u:2390xymu]
and glue them together in the square


[tex]\xymatrix{
F \ar[r]\ar[d] & \varnothing \ar[r]\ar[d]& 1\ar[d] \\
\varnothing \ar[r] & 1\ar[r] & C
}[/tex][/list:u:2390xymu]
It's easy to see that \(F\cong \varnothing\) and \(C\cong 1\amalg 1\) (this is true independently from the pullout axiom); so now the square


[tex]\xymatrix{
F\ar[r]\ar[d] & \varnothing \ar[d]\\
1 \ar[r]& C
}[/tex][/list:u:2390xymu]
must be a pullout, and \(F\to\varnothing\) can only be the identity arrow, so that the arrow \(1\to C\) is an isomorphism too (the class of isomorphisms is closed under cobase change). This entails that there is an isomorphism \(1\to\varnothing\cong 1\amalg 1\), which allows to conclude:


[tex]\xymatrix{
\varnothing \ar[r]\ar[d] & 1\ar[d]^{id}\\
1 \ar[r]& 1
}[/tex][/list:u:2390xymu]
is a pullout and isomorphisms are closed under base change too.

We now use this fact to show that products and coproducts coincide everywhere: build the outer square in the following diagram, out of the smaller square (which are precisely the pullbacks/pushouts needed to define products and coproducts):


[tex]\xymatrix{
Y\ar[r]\ar[d] & X\times Y\ar[r]\ar[d] & Y \ar[d] && X\ar[r]\ar[d] & X\times Y\ar[r]\ar[d] & X \ar[d]\\
0\ar[r]\ar[d] & X\ar[r]\ar[d] & 0\ar[d] && 0\ar[r]\ar[d] & Y\ar[r]\ar[d] & 0\ar[d]\\
Y\ar[r] & X\amalg Y\ar[r] & Y && X\ar[r] & X\amalg Y\ar[r] & X
}[/tex][/list:u:2390xymu]
These two diagrams imply the presence of biproducts, in the form of a result shown by Freyd (Abelian Categories, I don't remember the page): there exists an object \(S\) (the biproduct of \(X,Y\), hence unique up to unique isomorphism) such that


[*:2390xymu] There are arrows \(Y\leftrightarrows S\leftrightarrows X\);[/*:m:2390xymu]
[*:2390xymu] The arrow $Y\to S\to Y$ compose to the identity of $Y$, and the arrow $X\to S\to X$ compose to the identity of $X$;[/*:m:2390xymu]
[*:2390xymu] There are "exact sequences'' (in the sense of a pointed, finitely bicomplete category) $0\to Y\to S\to X\to 0$ and $0\to X\to S\to Y\to 0$.[/*:m:2390xymu][/list:u:2390xymu]
It is evident that the diagrams above contain all these informations.