# Relationship between pullbacks and the Ore condition

Let $C$ be a category and assume either that $C$ has all binary pullbacks or that $C$ satisfies right calculus of fractions. In both cases the localization of $C$ at every morphism (i.e. the groupoidification) can be represented by spans, i.e the objects are the same as in $C$ and morphisms are spans $A\leftarrow B\rightarrow C$ modulo some relation (see the link, section Construction of the localization).

My question is now: Does having binary pullbacks imply the right calculus of fractions? Or vice versa? Or is there no general relation between the two?

More generally: let $C$ have pullbacks and consider a class of morphisms $W$ containing the isos, pullback stable and satisfying 2 from 3. Benabou call this a pullback congruence. Lemma 1.2 of his paper "Some remarks on 2-categorical algebra" shows that such $W$ admits a calculus of right fractions.
Even better the category of fractions $C(W^{-1})$ admits pullbacks which are preserved by the quotient map $p:C \to C(W^{-1})$, and furthermore $p$ inverts precisely the $W$'s. This is all in the above paper.
• Nice, thank you! Just a little comment: Since then $W$ admits calculus of right fractions, all finite limits exist in $C(W^{1})$ and are preserved by the quotient map, provided they exist in $C$. The other property that $p$ inverts precisely the $W$'s is sometimes called saturated. – Werner Thumann Nov 29 '13 at 9:06
The question above has already been answered by john. But I want to answer here a question which may come to ones mind, when one reads the above answer. Namely the following: The original question is concerned with the special case $C=W$ if one considers right calculus of fractions for a pair $(C,W)$. Calculus of right fractions says that,
Define $C$ to be the category with objects natural numbers exluding $0$. An arrow from $n$ to $m$ is a sequence $(n_1,...,n_m)$ of $m$ natural numbers $\neq 0$ which sum up to $n$. Composition is as follows: Suppose you want to compose $(n_1,...,n_m):n\rightarrow m$ with $(m_1,...,m_k):m\rightarrow k$. To do this, divide the sequence $(n_1,...,n_m)$ into $k$ blocks of lengths $m_1,m_2,...,m_k$ and sum up the blocks, i.e. the composition is $$(n_1+...+n_{m_1},n_{m_1+1}+...+n_{m_2},...)$$ Property 1) is not hard to verify. But property 2) is not satisfied since $(2,1)$ and $(1,2)$ are coequalized by $(2)$ but cannot be equalized.