An exact (small) category $P$ is an environment in which we make sense of the "put-together"-edness of objects via (short) exact sequences. It seems like the K-theory of an exact category encodes the high order relations of how objects fit together, but I can't see how the $Q$-construction is the natural medium for this.

$\bullet$ $0 \to B \to A \to C \to 0$ means that $A$ is put together in some way from $B$ and $C$. Letting $E$ denote the category of short exact sequences in $P$ (with obvious morphisms), we see that this holds in the large as well: If $p$ (resp. $q$) is the projection functor $E \to P$ that sends a s.e.s. to the first (resp. last) object in the sequence, then $(K_i(p), K_i(q)): K_i(E) \to K_i(P) \oplus K_i(P)$ is an isomorphism.

$\bullet$ Moreover, results like devissage, localisation, and resolution are also easily seen to be reassurances that we are distilling a very sensible notion of put-together-edness.

Can anyone offer a (non-circular) reason as to why the Q-construction is right?

I don't see the sense or utility considering a category in which a morphism from $X$ to $Y$ is an isomorphism of $X$ with an (admissable) subquotient of $Y$.

  • $\begingroup$ math.mit.edu/~clarkbar/papers.html Check out the papers with the obviously related titles. $\endgroup$ – Dylan Wilson Apr 26 '13 at 6:32
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    $\begingroup$ I think that Waldhausen's S. construction answers your question, as it gives the intuition you're looking for. Waldhaunsen's S. is not Quillen's Q, but I read once that Quillen knew about S. since he defined Q, but he didn't like that S. was bisimplicial, so he looked for an equivalent, just simplicial construction, and so he found Q, which is 'less natural' also from my point of view. However I don't remember where I read this, nor I have any evidence of its veracity. $\endgroup$ – Fernando Muro Apr 26 '13 at 6:40

There's an interesting motivation in the paper by G. Segal:

K-homology theory and algebraic K-theory. K-theory and operator algebras (Proc. Conf., Univ. Georgia, Athens, Ga., 1975), pp. 113–127. Lecture Notes in Math., Vol. 575, Springer, Berlin, 1977

It is well-known that the space of Fredholm operators gives a model for topological K-theory $\Bbb Z \times BU$.

According to Segal, if $C$ is an exact category, a morphism of the Q-construction $QC$ is akin to a Fredholm operator.

So algebraic K-theory via the Q-construction mimics topological K-theory defined via Fredholm operators.

  • $\begingroup$ This is the sort of answer I was hoping someone had. Thanks! $\endgroup$ – Joshua Seaton Apr 26 '13 at 21:51

1) One wants the $Q$-construction to have the property that $\pi_1(QP)=K_0(P)$.

2) To get this, one wants, for any covering space of $BQP$, that $K_0(P)$ acts naturally on the fiber over $0$.

3) To get this, one wants to associate to any monomorphism $i$ in $P$ a morphism $i_!$ in $QP$ and to any epimorphism $j$ in $P$ a morphism $j^!$ in $QP$ in a way that satisfies certain simple properties; the statement of the properties, and the proof that they suffice to get this result, is in Quillen's Algebraic K-Theory I (Theorem I).

4) Quillen's $Q$-construction is the Universal construction yielding such $i_!$ and $j^!$. (The proof is in QUillen's paper, immediately preceding Theorem 1.)

5) Therefore, there's a sense in which Quillen's $Q$-construction is the natural first guess for what should work. (Of course the "naturality" of this guess appears only in hindsight; a lot of other people failed to find this construction.)

PS. After you work through the constructions, you see that this is another way to see the same thing: For any object $A$ in your category $P$, you want to associate the $K_0$-class $[A]$ to some loop in $BQP$. The simplest thing to hope for is two canonically defined maps from $0$ to $A$ in $QP$, which together give you your loop. Quillen's construction provides those two maps (recognizing $0$ as a quotient of both $0\subset A$ and $A\subset A$) in the simplest possible way.

  • $\begingroup$ Thank you very much for spelling this out. +1. It's nice to see how one can naturally reverse-engineer the Q-construction, which I have yet to see anywhere else. I am going with the other answer, as I was hoping for an answer less along the lines of it's right because we are naturally led to it' and more along the lines of here's why you shouldn't be surprised that we've ended up here. In any case, thanks again. $\endgroup$ – Joshua Seaton Apr 26 '13 at 21:50

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