Why are sheaves of a coverage the same as those on its generated Grothendieck coverage?

Question

Preliminaries: There are lots of variations on the settings in which we define sheaves. I am concerned with the details linking the general definition below to the Grothendieck topology it generates; particularly, the assertion that sheaves with respect to a coverage are the same as sheaves with respect to the generated Grothendieck coverage.

Let $\mathcal{C}$ be a (small) category. Do not assume $\mathcal{C}$ has pullbacks. Denote the Yoneda embedding of an object by $Yc$.

Descent can be phrased concisely in terms of subfunctors of Yoneda embeddings. We say, a presheaf $X$ satisfies descent with respect to a subfunctor $\iota \colon F \hookrightarrow Yc$ if and only if the precomposition $-\iota \colon \lbrack \mathcal{C}^{\operatorname{op}}, Set \rbrack (Yc,X) \rightarrow \lbrack \mathcal{C}^{\operatorname{op}}, Set \rbrack (F,X)$ is an isomorphism.

Any set of maps $f:=\{ f_i \colon c_i \rightarrow c \}_I$ generates a subfunctor $\iota_f \colon Y_f \hookrightarrow Yc$ by declaring $Y_f \bullet = \{ \alpha \colon \bullet \to c ~|~ \exists f_i ~ \alpha \mbox{ factors through } f_i \}$. In this manner, we can encode the usual "$X$ descends along the cover $f$" with "precomosition with $\iota_f$ is a isomorphism".

A site is a category equipped with a coverage $\mathcal{J}$ assigning to each object $c$ a collection of covering families $\mathcal{J}c$ which are sets of maps $f = \{ f_i \colon c_i \rightarrow c \}_I$ such that the covering families are supported under pullback i.e. $\forall f \in \mathcal{J}c,~\varphi \in \mathcal{C}(b,c)~ \exists g \in \mathcal{J}b: Y_g \hookrightarrow \varphi^*Y_f$ where $\varphi^*Y_f$ is the pullback of $Y_f \hookrightarrow Yc$ along composition with $\varphi$.

Then a sheaf with respect to the site $(\mathcal{C},\mathcal{J})$ is a presheaf $X \colon \mathcal{C}^{\operatorname{op}} \to Set$ satisfying descent for all the $Y_f$.

We can generate a Grothendieck coverage $\mathcal{G}$ from any coverage (for each $c$ intersect $\mathcal{G}_\lambda c$ all such coverages $\mathcal{G}_\lambda$ and note that the trivial coverage assigning all subsets of maps to $c$ as covering families is Grothendieck). What is not clear to me is how, "One can then show that for every coverage, there is a unique Grothendieck coverage having the same sheaves." as the nLab entry for coverage claims.

Question:

How is it that sheaves with respect to $\mathcal{G}$ are the same as those with respect to $\mathcal{J}$? In particular, given $X \in Sh(\mathcal{C},\mathcal{J})$, a cover $f \in \mathcal{J}$, and map $\varphi$; support under pullback gives us $g \in \mathcal{J}$ with:

$ Y_g \overset{\iota}{\hookrightarrow} \varphi^*Y_f \overset{\iota_{\varphi^*Y}}{\hookrightarrow} Yb$

That $X$ is a sheaf gives us precomposition with $\iota_g = \iota_{\varphi^*Y}\iota$ is an isomorphism. How does that imply precomposition with $\iota_{\varphi^*Y_f}$ is an isomorphism?

Answer 1

Extend $g$ to any generating set for $\varphi^*F$, say $\widehat{f} := \{ b_k \overset{\widehat{f}_k}{\longrightarrow} b \} \supseteq g := \{ b_{k_j} \overset{\widehat{f}_{k_j}}{\longrightarrow} b \}$, we will need to demonstrate that a sheaf with respect to the coverage $\mathcal{J}$ must satisfy descent for $\varphi^*F$.

Let $(y_k) \in \prod_K Xb_k$ be a $\varphi^*F$-compatible tuple. Because $X$ descends along $Y_g$, and $(y_{k_j})$ is $G$-compatible, $\exists!y \in Xb$ such that $X\widehat{f}_{k_j}y = y_{k_j}$. The problem can be restated as: there may be $\widehat{f}_k$ not in $g$ and a priori we don't know that $X\widehat{f}_k y = y_k$.

Let $\widehat{f}_t \in \widehat{f} - g$. Support under pullback of $Y_g$ along $\widehat{f}_t$ guarantees there is a cover $Y_h \in \mathcal{J} b_t$ generated by elements of $\widehat{f}_t^*G$ whence we have commuting squares $\widehat{f}_t h_s = \widehat{f}_{k_{j_s}}\psi_s$ with h = $\{ h_s \}$. In particular note $(X\psi_s y_{k_{j_s}})$ is $Y_h$-compatible and therefore applying descent, we may conclude $\exists!z \in Xb_t$ such that $Xh_sz = X\psi_s y_{k_{j_s}}$.

But by $Y_g$-compatibility $\widehat{f}_t h_s = \widehat{f}_{k_{j_s}}\psi_s$, we know $Xh_s y_t = X\psi_s y_{k_{j_s}}$ and thus $z = y_t$. In other words, $X$ satisfies descent for $\varphi^*F$ as desired.