# Finite tor dimension in Quillen's paper

Quillen gives the following projection formula his foundational paper on higher algebraic k-theory. (For simplicity, I assume all schemes are noetherian.)

Let $f: X \rightarrow Y$ be a proper map of finite tor dimension. Assume $X, Y$ have ample line bundles. Then, for $x \in K_0 X, y \in K_q' Y$ we have $$f_*(x \cdot f^* y) = f_* (x) \cdot y$$

Where he says $f_*: K_0 X \rightarrow K_0 Y$ is the map of SGA 6 2.12.2 (no expose given).

Question:

Notion of finite tor dimension

Quillen defines a morphism $f: X \rightarrow Y$ to be "of finite tor dimension" if $\mathcal{O}_X$ is of finite tor dimension as a module over $f^{-1}(\mathcal{O}_Y)$.

Suppose $X, Y$ are quasiprojective. Then, (ref: Fulton/MacPherson Categorical framework for the study of singular spaces) being a morphism of finite tor dimension is the same as being a perfect morphism in the sense of SGA 6.

$\textbf{Question}$: I wasn't able to deduce if Quillen's notion of finite tor dimension is the same as being perfect in the sense of the stacks project (say if $X,Y$ have ample line bundles). Is this true?

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Thanks for the comments so far. The question is still completely open. – abcd Jul 7 '13 at 5:15

I don't have enough rep to comment: usually the pushforward on grothendieck groups is defined using higher direct images (or derived categories is you prefer), so that $f_*[E] = \sum (-1)^i [R^if_* E]$.
For this to be well defined you need properness to still land in coherent sheaves and only finitely many $R^i$ (which is true for essentially any morphism of everything).
On the other hand you can define $f^*$ on the level of Grothendieck groups again by using the derived inverse image (which is essentially computing Tors for $f^{-1}O_Y$), here you use finite tor dimension.
by the way, what do you mean by $K'_q$? – John Salvatierrez Jul 4 '13 at 23:15
Thanks for the perspective. By $K'_q$ I mean Quillen's $K'$ theory, which agrees with $K-$theory on regular noetherian ... schemes. – abcd Jul 4 '13 at 23:20