Universal property of sheafification

If you have a map of presheaves $\varphi : \mathcal{F} \to \mathcal{G}$, this induces a map on stalks

$$ \varphi_p : \mathcal{F}_p \to \mathcal{G}_p $$

by the universal property of direct limits. Thus if we have an $f \in \mathcal{F}^+(U)$, we can compose $f$ with the map

$$ \bigsqcup_{p \in U} \varphi_p : \bigsqcup_{p \in U} \mathcal{F}_p \to \bigsqcup_{p \in U}\mathcal{G}_p $$

to get a map $U \to \bigsqcup_{p \in U} \mathcal{G}_p$. Call this $\tilde{\varphi}(f)$.Then we get a morphism $\tilde{\varphi}\colon\mathcal{F}^+\to \mathcal{G}^+$.

In fact, for every $p\in U$ there exists an open neighborhood $p \in V_p \subset U$ and $h \in \mathcal{F}(V_p)$ such that $h_q =f(q)$ for all $q \in V_p$ by definition of $\mathcal{F}^+$, and $\tilde{\varphi}(f)(q)=\varphi_q(f(q))=\varphi_q(h_q)=(\varphi(V_p)(h))_q$. So we can conclude $\tilde{\varphi}(f)\in \mathcal{G}^+(U)$.

Since $\mathcal{G}$ is assumed to be sheaf, then canonical morphism $\theta'\colon \mathcal{G}\to\mathcal{G}^+$ is isomorphic and so $\overline{\varphi}\colon={\theta'}^{-1}\circ\tilde{\varphi}$ gives a well defined morphism $\mathcal{F}^+ \to \mathcal{G}$.

Then we want $\overline{\varphi}\circ\theta=\varphi$.For this, take arbitary $x\in X$, then it is enough to see stalk at $x$; $\varphi_x={\theta'_x}^{-1}\circ\tilde{\varphi}_x\circ\theta_x$.This is easy to check.Uniqueness follows from the fact that $\theta_x$ is isomorphic for any $x\in X$.

uniqueness of $\overline{\varphi}$

Let $\psi\colon \mathcal{F}^{+}\to \mathcal{G}$ also satisfy $\psi\circ\theta=\varphi$.

Take stalk at for any $x\in X$; \begin{align} \overline{\varphi}_x\circ\theta_x=\varphi_x,\psi_x\circ\theta_x=\varphi_x. \end{align}

Since $\theta_x$ is isomorphic, \begin{align} \overline{\varphi}_x=\varphi_x\circ\theta_x^{-1}=\psi_x. \end{align} Therefore $\overline{\varphi}=\psi$.

existence of $\overline{\varphi}$

Take $s\in \mathcal{F}^+(U)$.By definition of $\mathcal{F}^+(U)$, for every $x\in U$ there exists an open neighborhood $x\in V_x\subset U$ and $t^x\in \mathcal{F}(V_x)$ such that $s(y)=(t^x)_y$ for all $y\in V_x$.

Then $\{V_x\}_{x\in U}$ is open covering of $U$ and $(\varphi(V_x)(t^x))_{x\in U}\in \prod_{x\in U}\mathcal{G}(V_x)$ satisfies gluing property.

In fact, for any $x,x'\in U$, first we get $t^x|_{V_x\cap V_{x'}}=t^{x'}|_{V_x\cap V_{x'}}$.

(because for arbitrary $y\in V_x\cap V_{x'}$,

\begin{align} (t^x|_{V_x\cap V_{x'}})_y=(t^x)_y=s(y)=(t^{x'})_y=(t^{x'}|_{V_x\cap V_{x'}})_y) \end{align}

So, \begin{align} \varphi(V_x)(t^x)|_{V_x\cap V_{x'}} =\varphi(V_x\cap V_{x'})(t^x|_{V_x\cap V_{x'}}) =\varphi(V_x\cap V_{x'})(t^{x'}|_{V_x\cap V_{x'}}) =\varphi(V_{x'})(t^{x'})|_{V_x\cap V_{x'}}. \end{align}

Therefore there uniquely exist $g\in \mathcal{G}(U)$ such that $g|_{V_x}=\varphi(V_x)(t^x)$ for all $x\in U$ since $\mathcal{G}$ is sheaf.Then we set $\overline{\varphi}(U)(s)\colon=g$.

What we have to do first is to check this $\overline{\varphi}$ is surely morphism of sheaf.

Let $V\subset U$ be open sets,then it is enough to show $\overline{\varphi}(U)(s)|_V=g|_V=\overline{\varphi}(V)(s|_V)$ under $\overline{\varphi}(U)(s)=g$.

First we can choose $V_x$ for $V'_x\colon=V_x\cap V$, $t^x$ for ${t'}^x \colon=t^x|_{V'_x}$ when we set $U=V,s=s|_V$.

(because for any $y\in V'_x$, \begin{align} s|_V(y)=s(y)=(t^x)_y=(t^x|_{V'_x})_y=({t'}^x)_y) \end{align}

Then \begin{align} (g|_V)|_{V'_x}=g|_{V'_x}=(g|_{V_x})|_{V'_x}=(\varphi_{V_x}(t^x))|_{V'_x}=\varphi_{V'_x}(t^x)|_{V'_x}=\varphi_{V'_x}({t'}^x). \end{align}

So $\overline{\varphi}(V)(s|_V)=g|_V$.This implies $\overline{\varphi}$ is morphism of sheaf.

Now only $\overline{\varphi}\circ\theta=\varphi$ is left to show.

Let $u\in \mathcal{F}(U)$.

When we set $U=U,s=\theta(U)(u)$, we can choose $V_x=U,t^x=u$ since for any $y\in U$, $s(y)=u_y$(by definition of $\theta$).

So, $\overline{\varphi}(U)(s)=\varphi(U)(u)$, that is, \begin{align} \varphi(U)(u)=\overline{\varphi}(U)(s)=\overline{\varphi}(U)(\theta(U)(u)). \end{align}