Let $\mathcal C$ be a category and suppose $\cal B \subseteq C$ is a full subcategory. Let $i \colon \mathcal B \longrightarrow \cal C$ denote the inclusion functor. Suppose that $S \subseteq \operatorname{Mor}\mathcal C$ is a class of morphisms in $\mathcal C$. Then we get a functor

$$\tilde i \colon \mathcal B[(S \cap \operatorname{Mor} \mathcal B) ^{-1}] \longrightarrow \mathcal C [S^{-1}]$$ If $S$ satifies the Ore conditions, and if for every morphism $s \colon M \longrightarrow N$ in $S$ with $M \in \mathcal B$ there exists a morphism $u \colon N \longrightarrow P$ such that $us \in S$ and $P \in \mathcal B$, then this functor $\tilde i$ is fully faithful.

Are there more general results that say when $\tilde i$ is fully faithful?

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    $\begingroup$ For $A, B\in \mathcal{B}$ every cospan $(s, X, f)$ (then $s: A\to X$, $f: B\to X$) in connected to a span $(s', X', f')$ with $X'\in \mathcal{B}$ (i.e. a sequence $X \to X_1 \leftarrow X_2 \to \ldots \leftarrow X_n\to X'$ where each $X_i$ is a top of a cospan and the arrows are in $S$ and are cospan morphisms) $\endgroup$ – Buschi Sergio Aug 28 '13 at 14:19
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    $\begingroup$ If you suppose the similar condiction for spans too, I seems that the Ore condiction (by the right cancellability) isn't neccessary. $\endgroup$ – Buschi Sergio Aug 28 '13 at 14:45
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    $\begingroup$ @BuschiSergio I'm sorry, I don't follow, is it possible to give a full description of what you are saying? $\endgroup$ – Paul Slevin Aug 28 '13 at 17:50

Let $\mathscr{C}$ a category and let $\Sigma \subset \mathscr{C}$ be a wide subcategory (i.e. closed under composition and containing identites).

There exists (generally in a more large sets universe) the category of fractions $P: \mathscr{C} \to \mathscr{C}(\Sigma )$, and $P$ is the identity map on objects. Given a morphism $\hat{f}\in \mathscr{C}(\Sigma )(A, B)$ we say that a sequence $A \xleftarrow{s_1} Y_1 \xrightarrow{f_1} X_2\xleftarrow{s_2} Y_2 \ldots \xleftarrow{s_n} Y_n \xrightarrow{f_n} B$ in $\mathscr{C}$ with $s_i \in \Sigma \ 1 \leq i\leq n$, represent $\hat{f} $ if $P(f_n)\circ P(s_n^{-1} )\circ \ldots P(s_1^{-1} )\circ P(f_1)= \hat{f} $. From the construction of $\mathscr{C}(\Sigma )$ any $\hat{f}$ is represented by some sequences.

1) Let $\Sigma$ with the right calculus of fraction.

Then any $\hat{f}$ has a short representation as a span like $A\xrightarrow{s} X \xleftarrow{f} B$, $s\in \Sigma$, we call it a $\Sigma$-span and indicate it as $(A, s, X, f, B)$ or simply as $(s, X, f)$.

We ask when two $\Sigma$-spans are equivalent i.e. represent the some morphisms in $\mathscr{C}(\Sigma )$.

The answere is that two $\Sigma$-spans $(A, s, X, f, B)$, $(A, t, Y, g, B)$ are equivalent iff exist a commutative diagram in $\mathscr{C}$:

$$ \begin{array}{ccccccc} E& \xrightarrow{c} & X & \xrightarrow{f} & B \\ \parallel&& \downarrow s & &\parallel \\ E &\xrightarrow{u}& A & & B \\ \parallel && \uparrow t & & \parallel\\ E& \xrightarrow[d]{} & Y & \xrightarrow{g} & B \\ \end{array}$$

with $u\in \Sigma$ and $c, d \in \mathscr{C}$. (the proof is very tedious, in Gabriel Zisman there isn't a true proof, omit not obvious details, and in H. Shubert there is only a incomplete proof)

In other words $\mathcal{C}(\Sigma )(A, B) \cong \varinjlim_{(X, s)\in \mathscr{C} _\Sigma (A)^{op}} (X, B) = \varinjlim\ (\ \mathscr{C}_\Sigma(A)^{op} \xrightarrow{\pi } \mathscr{C}^{op} \xrightarrow{[-, B]} Set\ ) $ where $\mathscr{C}_\Sigma (A)$ is the full subcategory of $A \downarrow \mathscr{C}$ with objects the arrow that belong to $\Sigma $ , observe that $\mathscr{C}_\Sigma(A)^{op}$ is filtered. Now consider the natural map $\mu : \mathscr{B}(\Sigma \cap \mathscr{B})(A, B) \to \mathscr{C}(\Sigma )(A, B) $ this is surjective iff for each $\Sigma $-span $(s, X, f)$ there is a equivalent $\Sigma $-span $(t, Y, g)$ with $Y \in \mathscr{B}$ (i.e. a diagram as above). And $\mu $ is injective iff given a diagram as above with $X, Y\in \mathscr{B}$ then exist a similar diagram with $E\in \mathscr{B}$. Of course $\mu $ is bijective if $\mathscr{B}_\Sigma (A)^{op} \to \mathscr{C}_\Sigma (A)^{op}$ is a final functor.

2) $\Sigma $ without "calculus of right fractions" hypothesis.

suppose that for $A\in \mathscr{B}$, $Y \in \mathscr{C}$ each span $A \xleftarrow{s} Y \xrightarrow{f} X$ is connected to a span like $ A \xleftarrow{s'} B \xrightarrow{f'} X $ with $B \in \mathscr{B}$, and each cospan $A \xrightarrow{f} X \xleftarrow{s} Y $ is connected to a cospan like $ A \xrightarrow{f'} B \xleftarrow{s'} Y $ with $B \in \mathscr{B}$.

I claim that the map $\mu : \mathscr{B}(\Sigma \cap \mathscr{B})(A, B) \to \mathscr{C}(\Sigma )(A, B) $ is surjective.

Given $\hat{f}: A \to B$, chose a representing sequences $A \xleftarrow{s_1} Y_1 \xrightarrow{f_1} X_1\xleftarrow{s_2} Y_2 \ldots \xleftarrow{s_n} Y_n \xrightarrow{f_n} B$,

Now the span $A \xleftarrow{s_1} Y_1 \xrightarrow{f_1} X_1$ is connected by a span like $A \xleftarrow{s'_1} B_1 \xrightarrow{f'_1} X_1$ with $B_1 \in \mathscr{B}$ and we replace this second to first, and consider the cospan $B_1 \xrightarrow{f'_1} X_1 \xleftarrow{s_2}$ this is connected to a cospan like $B_1 \xrightarrow{f''_1} A_1 \xleftarrow{Y_2}$ with $A_1 \in B$, and so on.. at the end we have a sequence $A \xleftarrow{s'_1} B_1 \xrightarrow{f''_1} A_1\xleftarrow{s'_2} B_2 \ldots \xleftarrow{s'_n} B_n \xrightarrow{f''_n} B$ in $\mathscr{B}$ such that mapped by $P$ in $\mathscr{C}( \Sigma )$ (and inverting all the $s'_i$ $1 \leq i\leq n$) obtain $\hat{f} $.

  • $\begingroup$ Does "connected" mean "equivalent " ? $\endgroup$ – Paul Slevin Aug 29 '13 at 11:58
  • $\begingroup$ Also, can we comment on injectivity of $\mu?$ $\endgroup$ – Paul Slevin Aug 29 '13 at 14:09
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    $\begingroup$ two $\Sigma$-spans are equivalent if (definition) represent the some morphisms in $\mathscr{C}(\Sigma )$. $\endgroup$ – Buschi Sergio Aug 29 '13 at 16:36

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