Reference for "It is enough to specify a sheaf on a basis"?
This is an excellent question and to tell the truth it is often handled in a cavalier fashion in the literature. This is a pity because it is a fundamental concept in algebraic geometry.
For example the structural sheaf $\mathcal O_X$ of an affine scheme $X=Spec(A)$ is defined by saying that over a basic open set $D(f)\subset X \;(f\in A)$ its value is $\Gamma(D(f),\mathcal O_X)=A_f$ and then relying on the mechanism of sheaves on a basis to extend this to a sheaf on $X$.
The same procedure is also followed in defining the quasi-coherent sheaf of modules $\tilde M$ on $X$ associated to the $A$-module $M$.
However there are happy exceptions on the net , like Lucien Szpiro's notes where sheaves on a basis of open sets are discussed in detail on pages 14-16.
You can also find a careful treatment in De Jong and collaborators' Stack Project , Chapter 6 "Sheaves on Spaces", section 30, "Bases and sheaves"
I've wondered about this myself recently and haven't been able to find any source for a general statement involving $\mathcal{C}$-valued sheaves. I'm sure I've simply not looked hard enough (I don't do AG, for instance), but for the record, I'll sketch out a proof along with some other things one can deduce. The verification that the extension is a sheaf is long and mostly a very, very tedious verification that everything really does commute.
One can formulate the notion of a $\mathcal{C}$-valued (pre)sheaf on a base $\mathscr{B}^{op}\subset \mathcal{O}(X)^{op}$ for any reasonably complete category $\mathcal{C}$. The idea is exactly as it works with sheaves, just find the arrow-theoretic expression of the set axioms. Doing so, one has this formulation:
A presheaf $\mathcal{F}$ on a base $\mathscr{B}$ is said to be a sheaf on $\mathscr{B}$ if the following holds. Given any covering $(U_{i})_{i\in I}$ by basis sets $U_i$ of a basis set $U$, and for each pair $(i,j)\in I\times I$, a family of coverings $(U_{ij,k})_{k\in I_{(i,j)}}$ of the intersection $U_i\cap U_j$, the diagram with the two sets of restrictions $\mathcal{F}(U_i)\to \mathcal{F}(U_{ijk})$ and $\mathcal{F}(U_i)\to \mathcal{F}(U_{jik})$ replacing the maps appearing in the usual equalizer condition for a sheaf is an equalizer.
With that said, here's what I believe to be true.
Let $\mathcal{C}$ be a complete category, $X$ a space and $\mathscr{B}^{op}\subset \mathcal{O}(X)^{op}$ a base for the topology. Let $i\colon \mathscr{B}^{op}\hookrightarrow \mathcal{O}(X)^{op}$ be the inclusion.
- There is a functor of restriction on (pre)sheaves $i^\ast\colon (P)Sh_{X}(\mathcal{C})\to (P)Sh_{\mathscr{B}}(\mathcal{C})$.
- If $\mathcal{F}$ is a sheaf on $X$, then $i^\ast \mathcal{F}=\mathcal{F}\circ i$ is a sheaf on $\mathscr{B}$.
- If $\mathcal{F}$ is a presheaf on $\mathscr B$, then there is a presheaf $\mathcal{F}^e\colon \mathcal{O}(X)^{op}\to\mathcal C$ defined by $\mathcal{F}^e(U)=\lim_{\mathscr{B} \ni B\subset U}\mathcal{F}(U)$ with the obvious induced morphisms as well as a natural isomorphism $\mathcal{F}\mathop{\to}\limits^{\eta} \mathcal{F}^e\circ i$.
- If $\mathcal{F}$ is a presheaf on $\mathscr{B}$, then the data $(\mathcal{F}^e,\eta)$ is terminal among all other extensions of $\mathcal{F}$ to a presheaf on $X$.
- Upon making choices, $(-)^e$ assembles into a functor. One deduces what it does on arrows by understanding $F^e\circ i\to G^e\circ i$ to be the unique arrow $\varphi\colon F\to G$ fitting into the data of $(F,\eta)$, $(G,\tau)$ and the morphism $f\colon F\to G$ as $i^\ast \varphi$. For any other choice, $(-)^e_2$, there is a natural iso $(-)^e\approx (-)^e_2$.
- If $\mathcal{F}$ is a sheaf on $\mathscr{B}$, then $\mathcal{F}^e$ is a sheaf on $X$.
- The functors $i^\ast$ and $(-)^e$ are left and right adjoint on presheaves. The adjunction restricts on sheaves to an adjoint equivalence $i^\ast_{Sh} : Sh_{X}(C)\cong Sh_{\mathscr{B}}(C) : (-)^e$.
I'll sketch 2. and 6. since everything else follows more-or-less in a straightforward fashion from these.
2.
The essential ingredient is to factor (for given $U,U_i,U_{ijk}\in \mathscr{B}$) the equalizer test diagram for a sheaf on the base $\mathscr{B}$ as the sheaf test diagram for $U,U_i,U_i\cap U_j$, followed by a single arrow $\prod \mathcal{F}(U_i\cap U_j)\to \prod_{i,j} \prod_k \mathcal{F}(U_{ijk})$. (It turns out this arrow will be monic, since $\mathcal{F}$ is a sheaf and $\mathcal{C}$ complete so that $\prod$ commutes with equalizers.)
6.
This is actually fairly delicate in my account of it. The hardest part is making a reduction.
Notation/Conventions
Here, we'll collect a few notations and a few potentially useful (?) reminders/observations.
For ease of typing, I'm going to stop writing $\mathcal{C}$, and put $C=\mathcal{C}$. Moreover, I'm going to put $F:=\mathcal{F}$ the functor on $\mathscr{B}$, and $F^e$ be the extension.
For any open set $U$, define $\mathscr{B}_U:=\{B\in \mathscr{B} : B\subset U\}$ and identify these all as subcategories of $\mathscr{B}$.
I'm going to casually conflate $\mathscr{B}$ and $\mathscr{B}^{op}$ for ease of typing.
Since we're working on the level of a single sheaf on $\mathscr{B}$, it is easier notationally to suppose (WLOG) that ${F}^e$ extends ${F}$—that is, ${F}^e\circ i={F}$. If we don't assume this, the notation becomes even worse and I don't want to deal with it.
For the presheaf $F^e$, the restrictions $F^e(U)\to F^e(V)$ are induced by the projections associated to the cones $F^e(V)\to \left. F\right|\mathscr{B}_{V}$. In particular, by our assumption, if $U\notin \mathscr{B}$, $B\in \mathscr{B}$, then the projection $F^e(U)\to F(B)$ is precisely the restriction.
Denote $res$ the restrictions for $F$ and $res^e$ the restrictions for $F^e$. Then $res^e_{UB}res_{BB'}=res^{e}_{UB'}$.
Given $U,U_i,U_{i}\cap U_j$, let $$\mathscr{B}_{I}=\{B\in \mathscr{B} : B\subset U_i \text{for some }i\in I\}\ldotp$$ It is clear that $\mathscr{B}_I$ is a basis for $U$ from the definitions.
Consider the sheaf test diagram for $F^e$ of this family with arrow $f\colon c\to \prod_{i}F(U_i)$. A little reflection shows that if we can show that we can produce the lift $\tilde{f}\colon c\to \mathcal{F}^e(U)$ if we can show $F^e(U)=\lim_{\mathscr{B}_{I}\ni B}F(B)$.
Claim 1: Given $F$ a presheaf on $\mathscr{B}$, $U$ open, $V\subset U$, and $\mathscr{B}_0\subset\mathscr{B}$ a subcollection forming a basis for $U$, $F^e(V)=\lim_{\mathscr{B}_0\ni B\subset V}F(B)$.
For now, let's assume the claim.
(Existence)
Then we have $F^e(U)=\lim_{\mathscr{B}_{I}\ni B}F(B)$. Hence, the obvious choice would be to define $\tilde{f}\colon c\to \lim_{\mathscr{B}_{I}\ni B}F(B)$ on components to be the corresponding components of $f$—this will induce the desired arrow because it defines a cone $c\to \left. F\right|\mathscr{B}_I$ as the components of $f$ were already a cone. More precisely, let $f_i$ be the components of $f$, and let $f_{i,V}\colon c\to F(V)$ be the components of $f_{i}$ inducing it to the limit. Since $f$ equalizes the sheaf test diagram, one observes that:
- For any basis sets $V\subset W$ in $\mathscr{B}_I$, $res_{VW}f_{i,V}=f_{i,W}$.
- For $V\subset U_i\cap U_j$, $f_{j,V}=f_{i,V}$. Hence, whenever $V\subset U_k$, $f_{k,V}=f_{i,V}$.
- If $V\subset U_i$ and $V\nsubseteq U_j$ for any other $j$, then there is only one $i$ with component $f_{i,V}$.
Using these observations and the observation it is clear that we get a cone from $c\to \left. F\right|\mathscr{B}_I$. Since the arrow to $\prod F(U_i)$ is componentwise the projections $F^e(U)\to F(U_i)$, which are induced by the canonical projections $F^e(U)\to F(B)$ for $B\subset U_i$, the composite with $\tilde{f}$ on the $(i,B\subset U_i)$-th component is $pr_{i,B}pr_{i}^e\circ \tilde{f}=f_{i,B}$, by definition. This shows existence.
(Uniqueness)
Claim 2: Under the assumption of claim 1, uniqueness is obvious. (Check components, &c. &c.)
This shows that $F^e$ is a sheaf if we assume claim 1.
Proof of Claim 1 in the applicable case above
We assume $V=U$ in this case. Note that there is an obvious arrow $f\colon F^e(U)\to \lim_{\mathscr{B}_I\ni B\subset U}F(B)$ induced by projection.
Remark: I'm confident it's slicker to prove that $\lim_{\mathscr{B}_I\ni B\subset U}F(B)$ satisfies the UP of $\lim_{\mathscr{B}\ni B\subset U}F(B)=F^e(U)$. But since it is not that hard to build an inverse $\varphi_U$ to $f$ on the components, this will be our strategy
Any basis set $V\in \mathscr{B}_{U}$ (N.B., $\mathscr{B}_U\supseteq \mathscr{B}_I$) can be covered by sets in $\mathscr{B}_I$, and any intersection of basis sets $B_1,B_2$ can by covered by basis sets in $\mathscr{B}_{I}$. Since $F$ is a sheaf on $\mathscr{B}$, there is the obvious equalizer diagram involving these. Do this for each basis set $V\in \mathscr{B}_U$. This gives us a family of equalizers indexed over the small discrete category of $\mathscr{B}_{U}$. Since $C$ is complete, we can stick these into one giant equalizer (imagine I could draw an equalizer)
$$\prod_{V\in \mathscr{B}_U}\mathcal{F}(V)\to\prod_{V\in \mathscr{B}_U}\prod_{\mathscr{B}_I\ni V_i\subset V}F(V_i)\to \prod_{V\in \mathscr{B}_U}\prod_{\mathscr{B}_I\ni V_{ijk}\subset V_i\cap V_j}F(V_{ijk})\ldotp$$
This follows by the usual adjointness argument, $\prod$ is right adjoint to the diagonal functor and hence preserves/commutes with limits.
Let $g_V\colon \lim_{\mathscr{B}_I\ni B\subset U}F(U)\to \prod_{\mathscr{B}\ni V_i\subset V}F(V_i)$ be the projections $pr_{V_i}$. Then $res_{V_i,V_{ijk}}pr_{V_i}=res_{V_j,V_{ijk}}pr_{V_j}$ since the projections off of the limit are necessarily a cone to $\left. F\right|\mathscr{B}_I$. Doing this for each basis set $\mathscr{B}\ni V\subset U$, this assembles into an arrow $$\lim_{\mathscr{B}_I\ni B\subset U}F(U)\to \prod_{V\in \mathscr{B}_U}\prod_{\mathscr{B}_I\ni V_i\subset V}F(V_i)$$ that gets equalized in our new equalizer (namely, it is componentwise $g_V$, and since the equalizer in question was obtained componentwise, this follows from what we just said). Call it $\varphi$ with components $\varphi_V$. Define $\varphi=f^{-1}$ on components as $f^{-1}_V=\varphi_V$. If we are given $W\subset V\in \mathscr{B}_I$, then $res_{VW}\varphi_V=\varphi_W$ follows by the uniqueness of the UP of the equalizer—the arrow $res_{VW}\varphi_V$ also fits into the equalizer diagram corresponding to $W$.
Note that the uniqueness part above tells us that whenever $V\in\mathscr{B}_I$, $\varphi_V=pr_V$.
To check that $f \varphi=id$, note that on components $V\in\mathscr{B}_I$, this is $$pr_V f\varphi= (pr_V f)\varphi=res_{UV}^{e}\varphi=\varphi_V=pr_V,$$ which means $f\varphi$ is on components just the projections and hence must be the identity. Conversely, for $V\in\mathscr{B}$, $$res_{UV}^e \varphi f=(res_{UV}^e \varphi) f=\varphi_V f\ldotp$$ But $\varphi_V f$ fits into the equalizer for $V$, where $F^e(U)\to \prod_{\mathscr{B}_I\ni V_i\subset V} F(V_i)$ are the restrictions (i.e., the projections). Hence, by uniqueness, $\varphi_V f=res_{UV}^e$. Hence, $\varphi f=id$.
Remark: As I mentioned, I'm confident there's an easier way to do claim 1 by verifying universal properties. One might also be able to make quick work of it by showing the inclusion $\mathscr{B}_I^{op}\hookrightarrow \mathscr{B}^{op}_U$ is final in a different way, perhaps by using a characterization of such functors.
It is given in Daniel Perrin's Algebraic Geometry, Chapter 3, Section 2. And by the way, it is a nice introductory text for algebraic geometry, which does not cover much scheme theory, but gives a definition of an abstract variety (using sheaves, like in Mumford's Red book).
Added: I just saw that Perrin leaves most of the details to the reader. For another proof, see Remark 2.6/Lemma 2.7 in Qing Liu's Algebraic Geometry and Arithmetic curves.