A morphism of sheaves of sets on B is a morphism of presheaves of sets

1. Bases and Sheaves

Let X be a topological space and B be a base for the topology on X. A presheaf of sets on B is an assignment to each U ∈ B a set F (U ) and to each inclusion V ⊂ U of elements of Ba map ρ^U_V : F (U ) → F (V ) such that whenever W ⊂ V ⊂ U in B, we have ρ^U_W = ρ^V_W◦ ρ^U_V. Elements of F (U ) are also called sections (of F over U ). The image of s ∈ F (U ) in F (V ) is denoted by s|_V whenever V ⊂ U.

The stalk of F of presheaf of sets on B at x is defined to be F_x = colim_{U ∈B, x∈U}F (U ).

A morphism ϕ : F → G of pre sheaves of sets on B is a rule which assigns to each element U ∈ B a map of sets ϕ_U : F (U ) → G(U ) compatible with restriction map.

A sheaf F of sets on B is a presheaf on B satisfying the following additional property:

Given any U ∈ B, and any covering U =S

i∈IU_i with U_i ∈ B and any coverings U_i∩ U_j = S

k∈IijU_ijkwith U_ijk∈ B, the following sheaf condition holds: For any collection of sections s_i∈ F (U_i) such that for i, j ∈ I and k ∈ I_ij,

si|_Uijk = sj|_Uijk,

there exists a unique section s ∈ F (U ) such that s_i= s|_Ui for all i ∈ I.

A morphism of sheaves of sets on B is a morphism of presheaves of sets.

In this note, we will show that any sheaf of sets on B can be extended uniquely to a sheaf of sets on X:

Theorem 1.1. Let X be a topological space and B be a base for the topology on X.

Suppose F is a sheaf of sets on B. There is a unique sheaf F^ext of sets on X such that F^ext(U ) = F (U ) for all U ∈ B compatible with the restriction maps.

Let us observe that F (U ) can be identity with a subset ofQ

x∈UF_x with certain property:

Lemma 1.1. Given any open subset U ∈ B, we consider the productQ

x∈UF_x. Let us show that F (U ) can be identified with a subset ofQ

x∈UF_x via the map F (U ) → Y

x∈U

F_x

by s 7→ {s_x}_x∈U with the property that for any x ∈ U, there exists V ∈ B and s ∈ F (V ) so that for all y ∈ V, sy = (V, s) in Fy.

Proof. Let U ∈ B. Define ρ : F (U ) →Q

x∈UF_x sending s to {sx}_x∈U. Claim ρ is injective.

Suppose ρ(s) = ρ(s⁰). Then sx= s⁰_x for all x ∈ U. Choose representatives (s, Vx) and (s⁰, V_x⁰) of s_x and s⁰_x respective. Then there exists W_x∈ B so that W_x ⊂ V_x∩ V_x⁰ and s|_Wx = s⁰|_Wx. Notice that W_x∩ U is open in X but may not be in B. Since B is a base, we can choose W_x⁰ in B with x ∈ W_x⁰ so that W_x⁰ ⊂ W_x∩ U. Since W_x⁰ ⊂ W_x and s|Wx = s⁰|_Wx, s|_Wx⁰ = s⁰|_W⁰

x. Therefore for each x ∈ U, we can choose W_x⁰ ∈ B with W_x⁰ ⊂ U and s|_W⁰

x = s⁰|_W⁰

x. Hence we may write Wx for W_x⁰. Notice that {Wx} forms an open covering for U. Write t_x = s|Wx

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and t⁰_x = s⁰|_Wx. Then tx, t⁰_x ∈ F (W_x) for all x ∈ U with tx = t⁰_x. For each x, y, write W_x∩ W_y =S

kW_x,y,k with W_x,y,k∈ B. Since t_x = s|_Wx, we find tx|_Wx,y,k = ty|_Wx,y,k = s|W_x,y,k.

The fact that s⁰|_Wx = t_x and s|_Wx = t_x in F (W_x) for all x ∈ U and the sheaf property (uniqueness of sections), we find s⁰ = s.

Let {sx}_x∈U be an element of Q

x∈UF (U ) obeying the property mentioned in the state- ment of this proposition. For x ∈ U, choose V^x ∈ B with x ∈ V^x and σ^x in F (V^x) so that sy is the equivalent class of (σ^x, V^x) for all y ∈ V^x. In fact, such V^x can be chosen so that V^x ⊂ U. Then {V^x : x ∈ U } forms an open covering of U. For z ∈ V^x∩ V^y, (σ^x, V^x) is equivalent to (σ^y, V^y) at z because (σ^x, V^x) and (σ^y, V^y) are both equivalent to s_z by definition. Hence there exists Wx,y,z ∈ B with W_x,y,z ⊂ V^x∩ V^y and z ∈ Wx,y,z so that

σ^x|_Wx,y,z = σ^y|_Wx,y,z.

Notice that {W_x,y,z : z ∈ V^x∩ V^y} forms an open covering of V^x∩ V^y. By the property of sheaf of sets, there exists a unique s ∈ F (U ) so that s|_V^x = σ^x. It follows from the definition that the image of s in Fx is exactly sx. Hence ρ(s) = {sx}_x∈U. Thus an element {sx}_x∈U obeying the property given above lies in the image Im ρ of ρ.

If {sx}_x∈X ∈ Im ρ, then there is s ∈ F (U ) so that the image of s in F_x is sx. Let x ∈ U.

Since B forms a base for the topology of X, we can choose V ∈ B so that x ∈ V and V ⊂ U. Let σ be the restriction of s to V, i.e. σ = s|V in F (V ). By the commutativity of the diagram

F (U ) −−−−→ F_y

ρ^U_V





y ^id



 y F (V ) −−−−→ F_y

we have sy = (σ, V ) for all y ∈ V. 

Let us go back to the proof of Theorem 1.1.

Let us construct a sheaf of sets F^ext on X as follows.

Let U be “any” open subset of X and F^ext(U ) the subset of Q

x∈UF_x consisting of all {s_x}_x∈U with the property that for any x ∈ U, there exist V ∈ B with x ∈ V and a section σ ∈ F (V ) so that for all y ∈ V, sy = (σ, V ) in Fy. For any inclusion of open sets V ⊂ U, we define r_V^U : F^ext(U ) → F^ext(V ) by sending {s_x}_x∈U to {s_x}_x∈V. Then F^ext is a presheaf of sets on X so that for any U ∈ B, the map ρ : F (U ) → F^ext(U ) is an one-to-one correspondence by Lemma 1.1. Hence for any U ∈ B, we identify F (U ) with F^ext(U ) and say F^ext(U ) = F (U ). Now, let us prove that the presheaf of sets F^ext is indeed a sheaf on X.

Let {Ui} be any open covering of U where U is an open subset of X (not necessarily in B) and {U_i} be an open covering of U. For each i, let s_i∈ F^ext(U_i) be a section so that (1.1) si|_Ui∩U_j = sj|_Ui∩U_j, i, j.

Claim: there is a unique s ∈ F^ext(U ) so that s|Ui = si for all i.

Write si = {si,x}_x∈Ui for each i. Let us define

sx= si,x, for x ∈ Ui.

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Equation 1.1 says that si,x= sj,xfor all x ∈ Ui∩ U_j. Hence s = {sx}_x∈U gives a well-defined element of Q

x∈UF_x. By definition, s|_Ui = {s_x}_x∈Ui = s_i. If t = {t_x}_x∈U so that t|_Ui = s_i for all i, then tx = si,x for all x ∈ Ui, for all i, and thus t = s. Hence there is a unique s ∈Q

x∈UF_x so that s|_Ui = s_i in F (U_i). Now, claim s ∈ F^ext(U ).

Let x ∈ U. Then x ∈ Ui for some i. Then s|Ui = si ∈ F^ext(Ui). By definition, there exists V ∈ B with x ∈ V and a section σ ∈ F (V ) such that for all y ∈ V, s_y = s_i,y = (σ, V ) in F_y. By definition, s ∈ F^ext(U ).