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?


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Nov 29, 2013 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,
1) spans can be completed to commutative squares, and
2) parallel arrows, which are coequalized by some arrow also get equalized by some other arrow.
Of course existence of pullbacks implies 1). But does 1) imply 2)? Here is a counterexample:

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.


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