# Pseudo-filtered category

The following definitions are borrowed from Grothendieck's SGA 4, p.14-15.

Definition 1 A category $I$ is called pseudo-filtered if it satisfies the following conditions.

1) For every two morphisms with common domain $f\colon i \rightarrow j$ and $g\colon i \rightarrow j'$, there exists an object $k$ and two morphisms $u\colon j \rightarrow k$ and $v\colon j' \rightarrow k$ such that $u\circ f = v\circ g$.

2) For every two parallel morphisms $u,v\colon i\rightarrow j$ in $I$, there exists an object $k$ and an morphism $w\colon j\rightarrow k$ such that $w\circ u = w\circ v$.

Definition 2 A category $I$ is called filtered if it is pseudo-filtered, nonempty and connected.

My question Is the above definition of filtered categories equivalent to the usual one? If yes, how do you prove it?

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Nonempty is given. Wikipedia's condition 1 gives connectivity. Condition 2 is word for word. Can you prove the conditions 1 can be deduced from each other? – Rachmaninoff Mar 29 '14 at 13:42
@Rachmaninoff Dear Rachmaninoff, that's the point of the question. – Makoto Kato Mar 29 '14 at 14:06
Sure. Just wanted to simplify things. Basically, show given Groths dfn then for any pair of objects there are two arrows with respective domains and with common codomain. Conversely, given Wiki dfn, show 1 above. – Rachmaninoff Mar 29 '14 at 14:52
Wiki implies Groth is easy. Given $f\colon i\to j$ and $f'\colon i \to j'$ use Wiki1 to get $h\colon j \to k'$ and $h'\colon j'\to k'$. Then apply Wiki2 to $h'f'$ and $hf$ to complete Groth1. – Rachmaninoff Mar 29 '14 at 15:06
So now the point of the question is Groth implies Wiki1 – Rachmaninoff Mar 29 '14 at 15:07

It is clear that a filtered category is a nonempty connected pseudo-filtered category. We will prove the converse. Let $I$ be a nonempty connected pseudo-filtered category. Let $i, j$ be objects of $I$. Since $I$ is connected, there exist $n$ arrows connecting $i$ and $j$. It suffices to prove that there exist arrows $i \rightarrow k$ and $j \rightarrow k$. It is easy to prove this using induction on $n$ and the condition 1).