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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?
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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)

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  • $\begingroup$ 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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