HW2

Example I. $X,Y$ topological spaces, $G(U)=$ continuous functions $U\to Y$ which have relatively compact image. This is a subpresheaf of $F$ with $F(U)=\{f:U\to Y\text{mbox}iscontinuous\}$, but clearly need not be a sheaf.

\begin{proof} Define $\phi :G\to F$ such that $\phi (U):G(U)\to F(U),f\mapsto f$. It is easy to verify the following diagram commutes.

$$\text{require}AMScd\begin{array}{ccc}{\mathcal{F}}_1(U)& \stackrel{\phi (U)}{\to }& {\mathcal{F}}_2(U)\\ {\mathrm{res}}_{U,V}\downarrow & & \downarrow {\mathrm{res}}_{U,V}\\ {\mathcal{F}}_1(V)& \underset{\phi (V)}{\to }& {\mathcal{F}}_2(V)\end{array}$$

Hence, $\phi$ is a presheaf map. It remains to show for any sheaf $\mathcal{H}$ and $h:G\to \mathcal{H}$, there exists unique $\theta :F\to \mathcal{H}$ such that $h=\theta \circ \phi$.

We first prove the existence. Here assume that at least one of $X$ and $Y$ is locally compact. Since $f$ is continuous, for each $x\in U$ there exists $U_x$ such that $f{|}_{U_x}\in G(U_i)$. Thus, we may take $U={\cup }_{i\in I}{U}_i$ such that $f{|}_{U_i}\in G(U_i)$. Then $h(f{|}_{U_i})\in \mathcal{H}(U_i)$. Notice that $h(f{|}_{U_i}){|}_{U_j}=h(f{|}_{U_i\cap U_j})=h(f{|}_{U_j}){|}_{U_i}$ for any $i,j$. Since $\mathcal{H}$ is a sheaf, there exists $h\in H$ such that $h{|}_{U_i}=h(f{|}_{U_i})$ and then we define $\theta (f)=h$.

Now we prove the uniqueness. If ${\theta }_1,{\theta }_2:F\to \mathcal{H}$ are two distinct maps such that $h={\theta }_i\circ \phi$, then there exists $f$ such that $h_1={\theta }_1(f)$ and $h_2={\theta }_2(f)$ are different. For any ${\cup }_{j\in J}{U}_j=U$, there is $h_1{|}_{U_j}=h_1(f{|}_{U_j})=h_2(f{|}_{U_j})=h_2{|}_{U_j}$ and then $h_1=h_2$ by $\mathcal{H}$ being a sheaf, which is a contradiction.

Therefore, $F(U)=\{f:U\to Y\text{mbox}iscontinuous\}$ is a sheafification of $G$. Furthermore, $G$ is not a sheaf. For example, take $X=Y=\mathbb{R}$ and $U_i=(-i,i)$. Then there does not exist $f$ such that $f{|}_{U_i}=\mathrm{id}{|}_{(-i,i)}$. \end{proof}

Example J. $X$ a topological space, $F(U)=$ the vector space of locally constant real-valued functions on $U$, modulo the constant functions on $U$. This is clearly a presheaf. But every $s\in F(U)$ goes to zero in $\Pi F\left(U_i\right)$ for some covering $\left\{U_i\right\}$, while if $U$ is not connected, $F(U)\ne (0)$. Therefore it is not a sheaf.

\begin{proof} For every $s\in F(U)$, there exists $\cup U_i=U$ such that $s{|}_{U_i}$ is constant. If $F$ is a sheaf, then $s=\overline{0}\in F(U)$ and so $F(U)=\{0\}$. However, if $U=O_1\cup O_2$ is not connected, define $f:U\to \mathbb{R}$ such that $f{|}_{O_1}=-1$ and $f{|}_{O_2}=-1$, then $\overline{f}\in F(U)$ is non-zero, which is a contradiction. \end{proof}