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 :

  1. 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 $s_i$.
  2. Can we construct an example where this condition fails and Is the sheafification of Image presheaf somehow related to adding this extra condition?

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)

  • $\begingroup$ Thank you. I have asked some other questions as well. See if you can answer. $\endgroup$ – user87543 Feb 13 '17 at 5:00
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    $\begingroup$ @PraphullaKoushik The problem with your extra conditions is that they are (almost) always false. Indeed, there might be many choices for the $t_i$ (as preimage of the $s_i$), and with different choices, they not always glue (even if $s$ do belongs to $\operatorname{Im}(\psi_U)$ !). The only case where this issue doesn't come up is when there is only one preimage, that is when $\psi$ in into. But in that case, it is obvious that the image presheaf is a sheaf (it is isomorphic to $\mathcal{F}$). $\endgroup$ – Roland Feb 13 '17 at 11:45
  • $\begingroup$ This is better be a comment on the question and not here... I now realize that i was expecting more which is almost always false. $\endgroup$ – user87543 Feb 13 '17 at 11:52

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