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dependent_product_functor [2015/12/22 17:58] nikolaj |
dependent_product_functor [2019/09/28 18:11] nikolaj |
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| \end{CD} | \end{CD} | ||
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| - | Essentially, we get the section space $\prod_X p$ that we are after, if we just replace the projection $\pi_2:X\times Y\to X$ with the more general $p$. A section of this space must in general then not map an $x$ in $X$ to $x$ in $Y\times X$, but instead merely map it somehow into $p^{-1}(x)$. | + | Essentially, we get the section space $\prod_X p$ that we are after, if we just replace the projection $\pi_2:X\times Y\to X$ with the more general $p$. The function space we obtain need the not just be $Y^X$, but can be some more general so called section space. A section of this space must in general then not map an $x$ in $X$ to $x$ in $Y\times X$, but instead merely map it somehow into $p^{-1}(x)$. |
| So for an auxiliary object $A$, consider the projection $\pi_2:(A\times X)\to X$, formally defined as the arrow obtained by pulling back $!_A:A\to*$ along $!_X:X\to*$, i.e. $\pi_2:=!_X{}^*!_A$. This definition is appearent from the commuting square defining the product: | So for an auxiliary object $A$, consider the projection $\pi_2:(A\times X)\to X$, formally defined as the arrow obtained by pulling back $!_A:A\to*$ along $!_X:X\to*$, i.e. $\pi_2:=!_X{}^*!_A$. This definition is appearent from the commuting square defining the product: | ||
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| The point is that an arrow $\theta:\pi_2\to p$ in ${\bf{C}}/X$ exactly fulfills our condition by definition. A right adjoint $\prod_{!_X}$ to the pullback functor $!_X^*$ is defined as an isomorphism | The point is that an arrow $\theta:\pi_2\to p$ in ${\bf{C}}/X$ exactly fulfills our condition by definition. A right adjoint $\prod_{!_X}$ to the pullback functor $!_X^*$ is defined as an isomorphism | ||
| - | $${\bf{C}}/X[!_X{}^*!_A,p]\cong{\bf{C}}/X[!_A,\prod_{!_X}p]$$ | + | $${\bf{C}}/X[!_X{}^*!_A,p]\cong{\bf{C}}/*[!_A,\prod_{!_X}p]$$ |
| I.e. this provides an isomorphism of the above triangle to | I.e. this provides an isomorphism of the above triangle to | ||
