Let $\varphi: \mathcal{F}\rightarrow \mathcal{G}$ be a morphism of sheaves, $\mathcal{F},\mathcal{G}$ are sheaves of abelian groups. For open $U\subseteq X$, define $\text{Im} (\varphi)(U)=\text{Im}(\varphi(U))$. I would like to understand why this is not a sheaf.
Let $V\subseteq U$. Restriction map is given by $\text{Im}(\varphi)(U)\rightarrow \text{Im}(\varphi)(V)$ by restriction.
$s\in \text{Im}(\varphi(U))\subseteq \mathcal{G}(U)$ mapsto $s|_V$. It is easy to see that $s|_V\in Im(\varphi)(V)$.
$\xymatrix{\mathcal{F}(U) \ar[r]^{\varphi(U)}&\mathcal{G}(U)\\
\mathcal{F}(V) \ar[u]^{\text{Res}}\ar[r]^{\varphi(V)}&\mathcal{G}(V) \ar[u]^{\text{Res}}}$
Some how I am not able to draw this commutative diagram. Latex code is correct but here it is not working. Feel free to change this. It is not talking & in the code.
Let $x\in \text{Im} (\varphi(U))$ there exists $y\in \mathcal{F}(U)$ such that $x=\varphi(U)(y)$. We have
$$(\varphi(V)\circ \rm{Res})(y)=(\rm{Res}~\circ \varphi(U))(y)$$ i.e., $\varphi(V)(y|_V)=\psi(U)(y)|_V=x|_V$. So, $x|_V\in Im(\varphi(V))$. So, we have a well defined map $\text{Im} (\varphi(U))\rightarrow \text{Im} (\varphi(V))$ given by $x\mapsto x|_V$. commutativity of Restriction map follows from commutativity of restriction map in $\mathcal{G}$. Restriction map from an open set to itself is Identity map. So, $\text{Im}(\varphi)$ is a presheaf.
Let $U=\bigcup U_i$ and $s,t\in \text{Im}(\varphi)(U)$ such that $s|_{U_i}=t|_{U_i}$ for all $i$. As $\mathcal{G}$ is a sheaf and $s,t$ can be seen as sections of $\mathcal{G}(U)$ with equal restrictions on each $U_i$ implies $s=t$.
Let $U=\bigcup U_i$ and $s_i\in \text{Im}(\varphi)(U_i)$ such that $s_i|{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ for all $i,j$. As $\mathcal{G}$ is a sheaf and $s_i$ can be seen as sections over $\mathcal{G}(U_i)$ with restriction conditions as above, there exists $s\in \mathcal{G}(U)$ such that $s|_{U_i}=s_i$ for all $i$.
But the problem is to prove that $s$ is in $\text{Im}(\varphi)(U)$. As $s_i\in \text{Im}(\varphi)(U_i)$ there exists $t_i\in \mathcal{F}(U_i)$ such that $\varphi(U_i)(t_i)=s_i$.
Suppose, $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ for all $i,j$ imply that $t_i|_{U_i\cap U_j}=t_j|_{U_i\cap U_j}$ then, as $\mathcal{F}$ is a sheaf, there exists $t\in \mathcal{F}(U)$ such that $t|_{U_i}=t_i$ for all $i$.
We have $$\varphi(U)(t)|_{U_i}=(\rm{Res}~\circ \varphi(U))(t)=(\varphi(U_i)\circ \rm{Res})(t)=\varphi(U_i)(t|_{U_i})=\varphi(U_i)(t_i)=s_i.$$
So, we have $\varphi(U)(t)|_{U_i}=s_i=s|_{U_i}$ for all $i$. By identity axiom of sheaf, we then have $\varphi(U)(t)=s$ i.e., $s\in \text{Im}(\varphi)(U)$ and we are done.
My questions are :
- Is this justification prove that Image assignment is actually sheaf with the extra assumption that I made that restrictions of $t_i$ have same condition as that of restrictions of $s_i$.
- Can we construct an example where this condition fails and Is the sheafification of Image presheaf somehow related to adding this extra condition?
Best Answer
Yes there are examples. For an example,consider the sheaf of $\mathbb{C}^{\infty}$ functions in $S^1$ to $ \mathbb{\Omega}^1( S^1)$ by $ f \to df$. on the sections. Now see that $d \theta$ is defined on two open sets $S^1 - N ,S^1 - S$ where $N = (0,1), S= (0,-1)$. But the function on $S^1$ agreeing on these 2 open sets cannot be of the form $df$ .\ In other words, the first cohomology of $S^1$ is non-zero.( every closed form is not exact)