Lemma 8.2.1. This actually does give a presheaf.
Proof. Let $g : V'/U \to V/U$ be as above and similarly $g' : V''/U \to V'/U$ be morphisms in $\mathcal{C}/U$. So $f' = f \circ g$ and $f'' = f' \circ g' = f \circ g \circ g'$. Let $\phi \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y)$. Then we have
\begin{eqnarray*} & & (\alpha _{g \circ g', f})_ y^{-1} \circ (g \circ g')^\ast \phi \circ (\alpha _{g \circ g', f})_ x \\ & = & (\alpha _{g \circ g', f})_ y^{-1} \circ (\alpha _{g', g})_{f^*y}^{-1} \circ (g')^*g^\ast \phi \circ (\alpha _{g', g})_{f^*x} \circ (\alpha _{g \circ g', f})_ x \\ & = & (\alpha _{g', f'})_ y^{-1} \circ (g')^*(\alpha _{g, f})_ y^{-1} \circ (g')^* g^\ast \phi \circ (g')^*(\alpha _{g, f})_ x \circ (\alpha _{g', f'})_ x \\ & = & (\alpha _{g', f'})_ y^{-1} \circ (g')^*\Big( (\alpha _{g, f})_ y^{-1} \circ g^\ast \phi \circ (\alpha _{g, f})_ x \Big) \circ (\alpha _{g', f'})_ x \end{eqnarray*}
which is what we want, namely $\phi |_{V''} = (\phi |_{V'})|_{V''}$. The first equality holds because $\alpha _{g', g}$ is a transformation of functors, and hence
\[ \xymatrix{ (g \circ g')^*f^*x \ar[rr]_{(g \circ g')^\ast \phi } \ar[d]_{(\alpha _{g', g})_{f^*x}} & & (g \circ g')^*f^*y \ar[d]^{(\alpha _{g', g})_{f^*y}} \\ (g')^*g^*f^*x \ar[rr]^{(g')^*g^\ast \phi } & & (g')^*g^*f^*y } \]
commutes. The second equality holds because of property (d) of a pseudo functor since $f' = f \circ g$ (see Categories, Definition 4.29.5). The last equality follows from the fact that $(g')^*$ is a functor. $\square$
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