7
$\begingroup$

I'm referring (also for notations and terminology) to P. Johnstone, Sketches of an Elephant. A Topos Theory Compendium. Volume I. Clarendon Press. Oxford, 2002. The Lemma can be found at page 540. I can't manage to explicitly prove the following claim that Johnstone makes in his proof of the lemma:

Further, the compatibility of the $s_{ij}$ ensures that $(s_{i}\vert\ i\in I)$ is a compatible family relative to the $f_{i}$.

Here the $s_{ij}$ form a compatibile family for the compositions $(f_{i}h_{ij}\vert\ i\in I,\ j\in J_{i})$. If one fixes $i$, $(s_{ij}\vert\ j\in J_{i})$ trivially is a compatible family for the family $(h_{ij}\vert j\in J_{i})$ and since $A$ satisfies the sheaf axiom for all the families of the $h_{ij}$, there's a unique $s_{i}\in A(U_{i})$ such that $A(h_{ij})(s_{i})=s_{ij}$ for all $j\in J_{i}$. The problem now is the following: by the very definition of compatible family, I need to show that for each object $M$ of $C$ and for each couple of arrows in $C$, $p\colon M\rightarrow U_{i}$ and $q\colon M\rightarrow U_{\overline{i}}$ such that $f_{i}p=f_{\overline{i}}q$ (for some $i,\overline{i}\in I$), one has $A(p)(s_{i})=A(q)(s_{\overline{i}})$. I suppose I should work with the definition of the $s_{i}$ and the compatibility of $s_{ij}$ as suggested by the author, but I don't know how to do it, basically since there is no reason for p or q to factorize through some $h_{ij}\colon U_{ij}\rightarrow U_{i}$ or some $h_{\overline{i}\overline{j}}\colon U_{\overline{i}\overline{j}}\rightarrow U_{\overline{i}}$. So, how do I use compatibility of $s_{ij}$?

Thanks a lot.

$\endgroup$
  • $\begingroup$ It would be good if you also stated the lemma, or at the very least, what $s_{i,j}$ and $s_i$ are. $\endgroup$ – Zhen Lin Apr 11 '13 at 18:07
4
$\begingroup$

The claim in question is as follows (in my notation):

Let $\mathfrak{U}$ be a sink on $X$. Suppose $\mathscr{F}$ satisfies the sheaf condition for $U$ and, for each $f : U \to X$ in $\mathfrak{U}$, $\mathfrak{V}_f$ is a sink on $U$ such that $\mathscr{F}$ satisfies the sheaf condition for $\mathfrak{V}_f$. Then, $\mathscr{F}$ satisfies the sheaf condition for the family of composites $\mathfrak{W} = \{ f \circ g : f \in \mathfrak{U}, g \in \mathfrak{V}_f \}$.

This appears to be false without additional hypotheses. (I am embarrassed to say that I did not spot this last year when I studied this section.) Let us consider the following category $\mathcal{C}$: it has objects $U, V, W, X$ and morphisms $f : V \to U$, $g : W \to V$, $h : X \to V$. Take $\mathfrak{U} = \{ f, f \circ g, f \circ h \}$, $\mathfrak{V}_f = \{ g \}$, $\mathfrak{V}_{f \circ g} = \{ \textrm{id}_W \}$, $\mathfrak{V}_{f \circ h} = \{ \textrm{id}_X \}$; note that these are all sieves. Consider the trivial presheaf $\mathscr{F}$ such that $\mathscr{F} (Z) = \{ 0, 1 \}$ for all $Z$ in $\mathcal{C}$. Of course, this satisfies the sheaf condition for the above-mentioned sieves, but it does not satisfy the sheaf condition for $\mathfrak{W} = \{ f \circ g, f \circ h \}$: taking $s_{f \circ g} = 0$, $s_{f \circ h} = 1$ yields a matching family for $\mathfrak{W}$ (trivially), but there is no possible amalgamation for this matching family.


Here is one way of repairing the claim: we assume that $\mathfrak{U}$ and all the $\mathfrak{V}_f$ are covering sieves for a sifted coverage $T$, and that $\mathscr{F}$ is a $T$-sheaf. Now suppose $(s_k : k \in \mathfrak{W})$ is a matching family for $\mathfrak{W}$. By the sheaf condition, there exists a unique $t_f$ such that $t_f |_g = s_{f \circ g}$ for all $g$ in $\mathfrak{V}_f$. I claim $(t_f : f \in \mathfrak{U})$ is a matching family for $\mathfrak{U}$.

Indeed, suppose $\operatorname{dom} f = \operatorname{codom} h$; we must show $t_f |_h = t_{f \circ h}$. But, for each $l : Y \to \operatorname{dom} h$ in $\mathfrak{V}_{f \circ h}$, there is a $T$-covering sieve $\mathfrak{Y}$ on $Y$ such that $\{ h \circ l \circ m : m \in \mathfrak{Y} \} \subseteq \mathfrak{V}_f$, so $t_{f \circ h} |_{l \circ m} = s_{f \circ h \circ l \circ m} = t_f |_{h \circ l \circ m}$ for all $m$ in $\mathfrak{Y}$, and hence, $t_{f \circ h} |_l = t_f |_{h \circ l}$ by the sheaf condition for $\mathfrak{Y}$. Thus, $t_{f \circ h} = t_f |_h$, by the sheaf condition for $\mathfrak{V}_{f \circ h}$.

I suspect the condition that $T$ be a sifted coverage could be dropped, but I have not checked the details.

$\endgroup$
  • $\begingroup$ Zhen Lin, I've noticed you posted the claim of the lemma in your previous answer and I thank you for doing that: I just couldn't gain access to the net up to now...I'm running out of time at the moment: I'll try and understand your answer as soon as possible. Thanks! $\endgroup$ – Marco Vergura Apr 12 '13 at 7:09
  • $\begingroup$ Your counterexample seems to work properly...So do we really have to conclude Johnsotne made a mistake in his Elephant? $\endgroup$ – Marco Vergura Apr 14 '13 at 9:52
  • $\begingroup$ However, there's still something I can't undersand in your answer, when you try to repair the claim. How can you say that for each $l : Y \to \operatorname{dom} (h)$ in $\mathfrak{V}_{f \circ h}$, there is a T-covering sieve $\mathfrak{Y}$ on Y such that $\{ h \circ l \circ m : m \in \mathfrak{Y} \} \subseteq \mathfrak{V}_f$? From the Elephant, I only know that $\mathfrak{Y}\subseteq l^{*}(\mathfrak{V}_{f \circ h})=\{g\in Arr(C):\ codom(g)=Y\land l\circ g\in \mathfrak{V}_{f \circ h}\}$. $\endgroup$ – Marco Vergura Apr 14 '13 at 10:09
  • $\begingroup$ Moreover, I can't get why you can conclude that $t_{f \circ h} |_l = t_f |_{h \circ l}$ by the sheaf condition for $\mathfrak{Y}$. I apologize for my questions which will probably seem silly to you, but I really am a novice in this field of Mathematics. $\endgroup$ – Marco Vergura Apr 14 '13 at 10:35
  • $\begingroup$ @user72134 I know of at least one other mistake in this section, so it is not inconceivable. My claim about the existence of $\mathfrak{Y}$ is axiom (C) for coverages, and the sheaf condition for $\mathfrak{Y}$ says, among other things, that equality of elements of $\mathscr{F}(Y)$ can be detected by restricting along all the morphisms in $\mathfrak{Y}$. $\endgroup$ – Zhen Lin Apr 14 '13 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.