5
$\begingroup$

Consider the category $Space(G)$ of $G$-spaces. For $H\subset G$, there is a forgetful functor from $Space(G)$ to $Space(H)$. Also, for an object $X$ in $Space(G)$ and another object $X'$ in $Space(G')$, $X\times X'$ is in $Space(G\times G')$.

Consider also the category of $Mod(G)$ of modules over $H_G(pt)$. For $H\subset G$, there is a forgetful functor from $Mod(G)$ to $Mod(H)$. Also for a module $M$ in $Mod(G)$ and another module $M'$ in $Mod(G')$, $M\otimes_{\mathbb{Z}} M'$ is in $Mod(G\times G')$.

My question is whether there is an established terminology for the categories $C(G)$ parameterized by a group $G$ satisfying the axioms abstracting those listed above:

  • For $H\subset G$ there is a forgetful functor from $C(G)$ to $C(H)$
  • For $G$, $G'$ there is a "multiplication functor" from $C(G)\times C(G')$ to $C(G\times G')$

satisfying various relations. The equivariant cohomology functor $H_G: Space(G) \to Mod(G)$ is compatible with these two operations. Is there a terminology for such a functor?


Update: Pondering more about it, I think it is better to formulate the concept as "categories $\mathcal{C}$ contravariantly parameterized by another (symmetric) monoidal category $\mathcal{X}$ ", i.e.

  • for an object $X$ in $\mathcal{X}$, there is a category $\mathcal{C}(X)$
  • for a morphism $f:X\to Y$ between objects in $\mathcal{X}$, there is a functor $f:\mathcal{C}(Y)\to\mathcal{C}(X)$
  • for an object $X$ and $Y$ with its product $X\times Y$ in $\mathcal{X}$, there is a multiplication functor which defines, by $o_1\in \mathcal{C}(X)$ and $o_2\in\mathcal{C}(Y)$, an object $o_1\times o_2 \in \mathcal{C}(X\times Y)$.
$\endgroup$
1
  • 6
    $\begingroup$ Perhaps you're thinking of a lax monoidal pseudofunctor $\mathcal{X}^\textrm{op} \to \mathfrak{Cat}$. $\endgroup$
    – Zhen Lin
    Sep 14, 2012 at 4:43

1 Answer 1

7
$\begingroup$

Zhen is right that you can think of it as a lax monoidal pseudofunctor. In this paper, I called an equivalent structure a "monoidal fibration".

$\endgroup$
1
  • $\begingroup$ Thank you! I guess $H_G$ is then a functor between two monoidal fibrations... $\endgroup$ Sep 14, 2012 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.