Let $\mathcal{F}$ be a presheaf on $X$, let $U$ have open cover $\{U_i\}$ and define three maps:
$$\rho:\mathcal{F}(U)\to \prod_{i\in I} \mathcal{F}(U_i),\quad s\mapsto(s|_{U_i})_i$$ $$\sigma:\prod_{i\in I} \mathcal{F}(U_i)\to \prod_{(i,j)\in I^2}\mathcal{F}(U_i\cap U_j),\quad (s_i)_i\mapsto (s_i|_{U_i\cap U_j})_{(i,j)}$$ $$\sigma':\prod_{i\in I} \mathcal{F}(U_i)\to \prod_{(i,j)\in I^2}\mathcal{F}(U_i\cap U_j),\quad (s_i)_i\mapsto (s_j|_{U_i\cap U_j})_{(i,j)}$$
I wanted to play with this to find out what was going on with $\sigma,\sigma'$ so the smallest example is $U=U_1\cup U_2$.
Then $\rho(s)=(s|_{U_1},s|_{U_2})=(s_1,s_2)$ (latter is just fixing a notation)
Then $\sigma(s_1,s_2)=(s_1,s_1|_{U_2},s_2|_{U_1},s_2)$ and $\sigma'(s_1,s_2)=(s_1,s_2|_{U_1},s_1|_{U_2},s_2)$ Right?
Aren't the image of $\sigma$ and $\sigma'$ always just transpositions of one another?
Context: Algebraic Geometry I - Schemes with examples and exercises - Gortz:
He defines a presheaf to be a sheaf if the following sequence is exact:
$$\mathcal{F}(U)\stackrel{\rho}{\to}\prod_{i\in I} \mathcal{F}(U_i)\stackrel{\stackrel{\sigma}{\rightrightarrows}}{\sigma'}\prod_{(i,j)\in I^2} \mathcal{F}(U_i\cap U_j)$$
The ordering on $I^2$ that I gave I just picked myself. He says this exactness means that the image of $\rho$ is the set of elements $(s_i)_{i\in I}$ such that $\sigma((s_i)_i)=\sigma'((s_i)_i)$
If these are equal up to ordering, why would he mention this?(this being that exactness gives: $\sigma((s_i)_i)=\sigma'((s_i)_i)$)