Sheafification of a presheaf

Definition:  Let $\mathbf{F}$ and $X$ be topological space, and $\pi : \mathbf{F} \rightarrow X$ be a surjective map, then the triple  $( \mathbf{F}, \pi, X )$ is called an ètale space if $\pi$ is a local homeomorphism, i.e. for all point $p \in \mathbf{F}$, there exists an open neighborhood $W$ of $p$ such that $\pi(W)$ is open in $X$ and $\pi|_W : W \rightarrow \pi(W)$ is a homeomorphism. 

$\Omega$ will be used to denote the topology of $X$ and $\Omega_x$ for all the open subsets of $X$ containing the point $x \in X$.

For all $x \in X$, we say that $\mathbf{F}_x := \pi^{-1}(x)$ is a stalk of $\mathbf{F}$ at $x$. If $U \in \Omega$, $\Gamma(U, \mathbf{F})$ will denote the set of continuous maps  $\alpha: U \rightarrow \mathbf{F}$ such that $\pi \circ \alpha = id$. The elements of $\Gamma(U, \mathbf{F})$ will be called a section of $\mathbf{F}$ over $U$.

Let $\mathbf{F}(U) = \Gamma(U, \mathbf{F})$ for $U \in \Omega$. Because the section is just continuous functions, there exists a natural restriction map. Namely, if $V \subset U$ and $\alpha \in \mathbf{F}(U)$, then $res_{U,V}(\alpha) = \alpha|_V$, the function restriction. This turns $\mathbf{F}$ into a presheaf of sets.

Because sections are simply maps, we can easily check the sheaf axioms. Let $\{ U_i \}$ be an open cover of $U \in \Omega$. If $\alpha_1, \alpha_2$ agree on the covers, then 

$$\alpha_1(x) = \alpha_1|_{U_i}(x) = \alpha_2|_{U_i}(x) = \alpha_2$$

 as $x \in U$, so $x \in U_i$ for some $i$. 

Also, if $\alpha_i \in \mathbf{F}(U_i)$, then we can define $\alpha(x) = \alpha_i(x)$ where $U_i$ is an open cover containing $x$. If $U_j$ is another cover that contains $x$, then we have that it agrees on the intersection, so the above definition is well-defined. And by our construction $\alpha|_{U_i} = \alpha_i$ is clear.

Hence $\mathbf{F}$ is, in fact, a sheaf. To distinguish $\mathbf{F}$ which is a topological space, we will denote by $\mathbf{F}^b$ for the sheaf associated with the ètale space $\mathbf{F}$. 

Proposition: Let $( \mathbf{F}, \pi, X)$ be an ètale space, then there exists a natural bijection between the stalks $\mathbf{F}_x$ and  $\mathbf{F}^b_x$. 

Proof) Given $\alpha \in \Gamma(U, \mathbf{F})$ with $U \in \Omega_x$, we define $j_U : \Gamma(U, \mathbf{F}) \rightarrow \mathbf{F}_x$ defined by $\alpha \mapsto \alpha(x)$.  We have $\pi \circ \alpha = id \Rightarrow \pi(\alpha(x)) = x \Rightarrow \alpha(x) \in \pi^{-1}(x) = \mathbf{F}_x$.

Now let $\alpha_x \in \mathbf{F}^b_x$, then there exists $\alpha \in \Gamma(U, \mathbf{F})$ representing $\alpha_x$. We define a map

$$j_x : \mathbf{F}_x^b \rightarrow \mathbf{F}_x$$

defined by $j_x(\alpha_x) = \alpha(x)$. Given two $\alpha_1$ and $\alpha_2$ representing $\alpha_x$, call the open set that the two maps agree on by $U$, i.e. 

$$\alpha_1|_U = \alpha_2|_U$$

As $U$ must contain $x$, we see that $\alpha_1(x) = \alpha_1|_U(x) = \alpha_2|_U(x) = \alpha_2(x)$. Hence $j_x$ is a well-defined map.

Surjectivity: Let $p \in \mathbf{F}_x$. Then $p \in \mathbf{F}$. There exists a open neighborhood $W \ni p$ such that $\pi|_W : W \rightarrow \pi(W)$ is a homeomorphism. Let $\alpha = (\pi|_W)^{-1}$. We claim that $\alpha_x$, the image of $\alpha \in \Gamma(\pi(W), \mathbf{F})$ works.  

1) $\pi \circ (\pi|_W)^{-1}(q) = \pi(\pi^{-1}(q)) = q$ for all $q \in \pi(W)$, hence indeed $\alpha \in \Gamma(\pi(W), \mathbf{F})$. 

2) As $W$ was a neighborhood of $p$, it follows that $\alpha(x) = (\pi|_W)^{-1}(x) = p$. This is equivalent to saying that $j_x(\alpha_x) = p$.

Injectivity: Suppose that $j_x(\alpha_x) = j_x(\beta_x) = p$. By restricting the map to its intersection, we can find representatives of $\alpha_x$ and $\beta_x$ from same $\Gamma(U, \mathbf{F})$. Now pick $W$ a neighborhood of $p$ which induces a homeomorphism $\pi|_W : W \rightarrow \pi(W)$. We may assume that $W \subset U$ by replacing $W$ by $W \cap U$. We still get a homeomorphism because homeomorphisms are compatible with restrctions. We claim that $\alpha \sim (\pi|_W)^{-1}$. 

Pick a point $\pi(w) \in \pi(W)$. 

$$\pi \circ \alpha(w) = w$$

by definition of a section then

$$(\pi|_W)^{-1} \circ \pi \circ \alpha(w) = (\pi|_W)^{-1}(w)$$

Hence $\alpha$ and $(\pi|_W)^{-1}$ agrees on $\pi(W)$. Their images are the same on stalk. Similarly, $\beta$ also agrres on $\pi(W)$, hence its image is the same with the image of $(\pi|_W)^{-1}$. Namely, $\alpha_x = \beta_x$.  

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Now our last task is to construct an ètale space from a presheaf $\mathcal{F}$. We define

$$\mathcal{F}^+ := \coprod_{x \in X} \mathcal{F}_x$$

where $\coprod$ is disjoint union.

We also define

$$\mathcal{W}(\alpha, U) := \{ \alpha_x ~|~ x \in U \} \text{ where } \alpha \in \mathcal{F}(U) \text{ and } U \in \Omega_x$$

We claim that $\mathcal{W}(\alpha, U)$ form a base of a topology. Consider two such sets $\mathcal{W}(\alpha, U)$ and $\mathcal{W}(\beta, V)$. Let $p$ be a point in the intersection, then there exists $\alpha_x , \beta_x$ such that $\alpha_x = p = \beta_x$. Because they are the same germ, we must have $W \in \Omega_x$ such that $\alpha|_W = \beta|_W$. $W$ contains $x$ because it must be a subset of the intersection of $U$ and $V$. Let $\gamma = \alpha|_W = \beta|_W$, then $p \in \mathcal{W}(\gamma, W) \subset \mathcal{W}(\alpha, U) \cap \mathcal{W}(\beta, V)$. 

The topology on $\mathcal{F}^+$ is defined to be the topology generated by $\mathcal{W}(\alpha, U)$. Let $\pi : \mathcal{F}^+ \rightarrow X$ be the map where $\pi(p) = x$ if $p \in \mathcal{F}_x$. This is well-defined because we had a disjoint union. Pick $\alpha_x \in \mathcal{F}^+$, then we have a homeomorphism between $\mathcal{W}(\alpha, U)$ and $U = \pi(\mathcal{W})(\alpha, U)$. For any open subset $V \subset U$, we have that $\pi^{-1}(V) = \mathcal{W}(\alpha, V)$ which is also open. Any open subset of open set as a subspace topology is open in the original topological space. This proves that $\pi$ is a local homeomorphism.

Hence the triple $\mathcal{F}^+, \pi, X$ is an ètale space, and we denote the associated sheaf by $\mathcal{F}^a$ instead of $\mathcal{F}^{+b}$ for notational ease. 

For $\alpha \in \mathcal{F}(U)$, we associate $\alpha^+ \in \mathcal{F}^a(U)$ where $\alpha^+(x) = \alpha_x$ for all $x \in U$.  If  $\mathcal{F}$ is a presheaf of algebraic structures (e.g. groups, rings, $R$-modules, etc. ), we define the structure of $\mathcal{F}^a(U)$ in the obvious way such that the map $\alpha \mapsto \alpha^+$ is a homomorphism. 

Reference: [Iitaka] Algebraic Geometry (30-31)