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seperated_presheaf [2014/10/29 09:55] nikolaj |
seperated_presheaf [2014/10/29 09:55] nikolaj |
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| $s=t\implies \forall(x\in U).\,s(x)=t(x)$ | $s=t\implies \forall(x\in U).\,s(x)=t(x)$ | ||
| and as corollary we have | and as corollary we have | ||
| - | $s=t\implies \forall(V\subseteq U).\,s|_V=t|_V.$. | + | $s=t\implies \forall(V\subseteq U).\,s|_V=t|_V$ |
| + | . | ||
| Sections also fulfill function extensionality, which goes in the reverse direction | Sections also fulfill function extensionality, which goes in the reverse direction | ||
| Line 26: | Line 27: | ||
| and consequently, if $C_U$ is a covering of $U$, then | and consequently, if $C_U$ is a covering of $U$, then | ||
| $\left(\forall(V\in C_U).\,s|_V=t|_V\right)\implies s=t$ | $\left(\forall(V\in C_U).\,s|_V=t|_V\right)\implies s=t$ | ||
| + | . | ||
| Now if the $FU$'s of a sheaf are to be a sets of such sections over a topological space, then it should be that the maps $F(i)$ between them can be seen as restriction of the function domains. To this end, if $i:V\to U$ is the inclusion of a small open set $V$ in another $U$, require that if $s,t\in FU$ agree on all restricted domains which make up the cover of $U$, i.e. $F(i)(s)=F(i)(t)$, then there can be no other way in which they can differ, so $s=t$. | Now if the $FU$'s of a sheaf are to be a sets of such sections over a topological space, then it should be that the maps $F(i)$ between them can be seen as restriction of the function domains. To this end, if $i:V\to U$ is the inclusion of a small open set $V$ in another $U$, require that if $s,t\in FU$ agree on all restricted domains which make up the cover of $U$, i.e. $F(i)(s)=F(i)(t)$, then there can be no other way in which they can differ, so $s=t$. | ||
