
Warning! Enormous and possibly unreadable answer follows. But you should read it, since I believe it addresses your question. There's a section about halfway through where the actual answer starts.
$\def\sh{\mathcal}\def\on{\operatorname}\def\id{\mathrm{id}}$All your examples are, as others have described them, examples of "the" projection formula. In the context of the six functors, I know a very formal way of proving this isomorphism as a consequence of base change. Suppose you have a map $f \colon X \to Y$ of "spaces" (schemes, whatever) and you want to prove that for sheaves $\sh{F}$ and $\sh{G}$ on $X$ and $Y$ respectively, there is an isomorphism
$$f_!(\sh{F} \otimes f^* \sh{G}) \cong f_! \sh{F} \otimes \sh{G}.$$
First, you rewrite it in terms of the "external tensor product" $\sh{F} \boxtimes \sh{G}$ on $X \times Y$. This combination has three important properties:
If we have other maps $g \colon W \to X$ and $h \colon Z \to Y$, then on $W \times Z$: $$(g \times h)^* (\sh{F} \boxtimes \sh{G}) \cong g^* \sh{F} \boxtimes h^* \sh{G}.$$
Likewise, if the maps are $g \colon X \to W$ and $h \colon Y \to Z$, then on $W \times Z$ we have $$(g \times h)_! (\sh{F} \boxtimes \sh{G}) \cong g_! \sh{F} \boxtimes h_! \sh{G}.$$
Finally, if $X = Y$ and $\Delta_X \colon X \to X \times X$ is the diagonal map, then we have
$$\sh{F} \otimes \sh{G} \cong \Delta_X^* (\sh{F} \boxtimes \sh{G}).$$
It's best to think of property 3 as defining the usual tensor product, rather than the equation $$\sh{F} \boxtimes \sh{G} = \on{pr}_X^* \sh{F} \otimes \on{pr}_Y^* \sh{G}$$ defining the external tensor product. This is especially valid for representations: if $V$ is a representation of a group $G$ and $W$ of a group $H$, then the vector space $V \otimes W$ is naturally a representation of $G \times H$, even if $G = H$ when you are allowed to restrict to the diagonal and produce the usual tensor product representation.
Anyway, using this, you want an isomorphism:
$$f_! (\id, f)^* (\sh{F} \boxtimes \sh{G}) \cong \Delta_X^* (f \times \id)_! (\sh{F} \boxtimes \sh{G}).$$
(The lefthand side is obtained by writing $(\id, f) = (\id \times f) \Delta_Y$ and using property 1.) Obviously, it is more productive to just remove the sheaves and prove the natural isomorphism of functors. The combination $(\id, f) \colon X \to X \times Y$ is usually called $\Gamma_f$ and is the graph of $f$; it is the base change of the diagonal map $\Delta_X$:
$$\begin{matrix}
X & \xrightarrow{\Gamma_f} & X \times Y \\
{\scriptstyle f} \downarrow & & \downarrow{\scriptstyle\id \times f} \\
Y & \xrightarrow{\Delta_X} & Y \times Y
\end{matrix}$$
Therefore, it follows (effectively from proper base change, though it does not matter that $\Delta_Y$ is proper since we are using the $!$ pushforward) that
$$f_! \Gamma_f^* \cong \Delta_Y^* (f \times \id)_!$$
and that's the projection formula.
Connection with the examples
If you want to establish a projection formula in a more general context, you need the following ingredients:
Some kind of external tensor product (and, correspondingly, some kind of product of "spaces") satisfying point 3 above with respect to whatever functor you are calling "pullback".
Points 1 and 2 above for whatever you are calling "pullback" and "pushforward".
A base change isomorphism for fiber products of spaces.
Let me address these points onebyone for your examples.
Abelian sheaves and continuous pushforward/pullback. I am embarrassingly ignorant of "easy" examples of cohomology, which is to say that I do not actually know what is true of topological spaces. However, it is implied in Brian Conrad's notes (above Proposition 3.1) that any continuous map of topological spaces ringed by the constant sheaf $\mathbb{Z}$ is "flat" (well, it obviously satisfies the algebraic criterion anyway) and if that is the case, then the base change map is an isomorphism because of "flat base change" rather than proper. Points 1–3 are obviously satisfied here just as in the six functors formalism, so you get your projection formula.
Algebraic cycles and pullback/pushforward in the Chow ring. Take your "sheaves" to be closed subschemes in $X$ or $Y$, with pullback being flat base change and pushforward being proper schemetheoretic image. The "external tensor product" is just the obvious product of subschemes in $X \times Y$, whose restriction to the diagonal is of course the intersection, so the formalism of point 1 is satisfied. Note that the product in the Chow ring is generically the intersection product. The formalism of point 2 is also satisfied, since you can work independently with each coordinate. As for the basechange isomorphism, it is certainly true settheoretically, and I believe it will work schemetheoretically also since for affine schemes, the schemetheoretic image of a map $A \to B$ is the ring generated in $B$ by $A$, and this is stable under taking a tensor product with some extension of $A$. I admit I'm a little fatigued (I'm writing this paragraph last) so I am willing to accept criticism for being so vague.
Finite groups with induction (= pushforward)/restriction (= pullback). Here, obviously, the product of "spaces" is the product of groups and the external product of "sheaves" is the tensor product $V \otimes W$ of representations of two groups $G$ and $H$, considered in the natural way as a representation of $G$ and $H$. Point 1 is satisfied since $G$ and $H$ just act separately on each factor. Point 2 is obvious for restriction and for induction, we use the fact that darij gave in his answer: $\def\Ind{\operatorname{Ind}}\Ind_H^G$ is tensoring up with $k[G]$, and
$$(V \otimes_{k[G]} k[G']) \otimes_k (W \otimes_{k[H]} k[H']) = (V \otimes_k W) \otimes_{k[G \times H]} k[G' \times H'].$$
So point 2 is satisfied too. The base change isomorphism is tricky; it is not actually true in general that for subgroups $H, K \subset G$ we have $\Ind_H^G(V)_K \cong \Ind_{K \cap H}^K(V_{K \cap H})$, though we do have a map (the base change map, of course):
$$ \Ind_{K \cap H}^K(V _{K \cap H}) = V \otimes_{k [K \cap H]} k[K] \to V \otimes_{k[H]} k[G]_K = \Ind_H^G(V)_K$$
sending elements of $K \subset G$ into $G$, which is obviously $k[K \cap H]$linear. The two sides have dimensions (times $\dim V$), respectively, $[K : K \cap H]$ and $[G : H]$, where the former is also equal to $\#(K.H)/H$. So these indices are the same if and only if $G = K.H$, and this actually happens in the particular situation of the projection formula, where $K = G \subset G \times G$ and "$H$" is $G \times H \subset G \times G$. Then the base change map is an isomorphism by counting the (finite!) dimensions, and you get your projection formula again.
It is perhaps worth noting that the projection formula for group representations is very closely related to that for sheaves, since for any algebraic (for example, finite) group $G$, we can form the algebraic stack $*/G$, on which quasicoherent sheaves are identified with representations of $G$. So, modulo a satisfactory theory of base change for morphisms of stacks, this example is actually sort of the same as example 1.
Spaces with topological group actions and restriction (= pullback)/??? (= pushforward). Alas, I cannot comment on this example, since I do not understand modern (or even somewhat dated) homotopy theory. However, if I recall what I once knew about smash products, we do have for spaces $X$ and $Y$ over a space $A$ having distinguished sections from $A$ that (in sort of informal quotient notation)
$$X \wedge_A Y = X \times_A Y / (A \times_A Y = A, X \times_A A = A),$$
the "equalities" being that the $Y$coordinate or $X$coordinate is identified with whatever point of $A$ it is mapped to via the structure as $A$spaces. Therefore, if we have two base spaces $A$ and $B$, and spaces $X$ and $Y$ over them respectively, we have an obvious "external smash product" over $A \times B$ (I apologize for the hacky symbol):
$$X \mathop{\fbox{$\wedge$}} Y = X \times Y / (A \times Y = A \times B, X \times A = A \times B)$$
in the same way. And if $A = B$, this basechanges correctly to the diagonal copy of $A$, so at least the formalism of point 1 is satisfied, along with the restriction part of point 2. I can't say anything about the pushforward parts, alas, but perhaps you (if you are still following this at all!) can fill in the details now.


answered Jan 24 '12 at 1:49

