Sheaves on sites given by a (regular) cd-structure

Question

Let $C$ be a category equipped with a Grothendieck topology generated by a cd-structure (see https://ncatlab.org/nlab/show/cd-structure or Voevodsky's paper Homotopy theory of simplicial presheaves in completely decomposable topologies (JPAA version)). Voevodsky proves that if a cd-structure is regular, a condition which includes that the distinguished squares are pullback-squares, then a sheaf of sets $F$ on $C$ sends distinguished squares to pullback squares. My question is: are the other conditions in the definition of a regular cd-structure really necessary for this to be true?

It seems to me that, if a distinguished square $$ \begin{array}{ccc} A & \to & B \\ \downarrow & &\downarrow \\ C &\to & D \\ \end{array} $$ is a pullback square, and $F$ is a sheaf on $C$, then sections $b\in F(B)$ and $c\in F(C)$ which agree on $F(A)$, induce a unique matching family indexed by the covering sieve generated by $\{B\to D, C\to D\}$ (here one needs that the distinguished square is a pullback). Since $F$ is a sheaf, there is a unique amalgamation $d \in F(D)$ restricting to $b$ and $c$. This is equivalent to saying that the natural map $F(D) \to F(B)\times_{F(A)}F(C)$ is a bijection.

Is there something wrong with this argument? I would expect that, if this reasoning were correct, this would be mentioned somewhere, and not only as a consequence of the cd-structure being regular, which would be unnecessarily strong.

Answer 1

In general sections $b \in F(B)$ and $c \in F(C)$ that agree on $F(A)$ don't induce a matching family on $\{B \to D, C \to D\}$ though. The sheaf condition for that family is that $$ F(D) \to F(B) \times F(C) \rightrightarrows F(B \times_D B) \times F(B \times_D C) \times F(C \times_D B) \times F(C \times_D C) $$ is an equalizer. If $b$ and $c$ agree on $F(A) = F(B \times_D C) = F(C \times_D B)$ then their images are equal in the middle two factors, but there are also the maps $F(B) \rightrightarrows F(B \times_D B)$ and $F(C) \rightrightarrows F(C \times_D C)$ to consider.

If $B \to D$ is a monomorphism then the two maps $B \times_D B \to B$ are equal (we could regard them as the identity) and there is no problem. But, in general we could take any map $B \to D$ and set $A = C = \varnothing$ (let's assume we already know $F(\varnothing) = *$). That makes a pullback square, but the condition that $F$ is a sheaf for the family $\{B \to D\}$ is weaker than the statement that $F(D) \to F(B)$ is an isomorphism.

The regularity condition is basically what is needed to ensure that ignoring the $F(B \times_D B)$ and $F(C \times_D C)$ factors does not affect the sheaf condition.