The Unapologetic Mathematician

Mathematics for the interested outsider

Mappings Between Presheaves

As ever, we want our objects of study to be objects in some category, and presheaves (and sheaves) are no exception. But, luckily, this much is straightforward.

Remember that we ended up defining a presheaf as a functor. Given our topological space X we set up the partial order category \mathrm{Subset}(X), flipped it around to \mathrm{Subset}(X)^\mathrm{op} so the arrows pointed the opposite way, and then said a presheaf of sets is a functor \mathcal{F}:\mathrm{Subset}(X)^\mathrm{op}\to\mathbf{Set}. So the natural home for them is the functor category \mathbf{Set}^{\mathrm{Subset}(X)^\mathrm{op}}, where the morphisms are natural transformations.

So what does this mean for our usual case where we consider presheaves of sets, or of sets equipped with some algebraic structure? Well, it means that we map from one presheaf \mathcal{F} to another one \mathcal{G} by picking a map for each and every open set: \phi_U:\mathcal{F}(U)\to\mathcal{G}(U). But these maps must be compatible with the restrictions: if V\subseteq U then we must have \phi_V\circ\cdot\vert_V=\cdot\vert_V\circ\phi_U. That is, given an element in \mathcal{F}(U), we can either first restrict it to V and then map it by \phi_V to \mathcal{G}(V), or we can first map it by \phi_U to \mathcal{G}_U and then restrict the result to V. In either case, we should get the same answer.

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March 19, 2011 - Posted by | Topology

1 Comment »

  1. [...] Direct Image Functor So far our morphisms only let us compare presheaves and sheaves on a single topological space . In fact, we have a [...]

    Pingback by The Direct Image Functor « The Unapologetic Mathematician | March 21, 2011 | Reply


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