MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From a bisimplicial space $T$, one can consider the simplicial spaces $\underline p \mapsto T_{pp} $, $\underline p \mapsto |\underline q \mapsto T_{pq} |$, and $\underline q \mapsto |\underline p \mapsto T_{pq} |$, where $| \cdot |$ denotes geometric realisation. In a lemma (used in proving Theorems A and B, in 'Higher Algebraic K-theory I'), Quillen gives a proof that these three simplicial sets all have coincidental realisations, but I'm having touble understanding the reasoning.

The idea of the proof is to first prove it in the special case of when $T$ is of the form $h^{rs} \times S$, where $h^{rs} = \hom_{\Delta}( -, \underline{r}) \times \hom_{\Delta}( -, \underline{s})$ (with the hom-sets having the discrete topology) and $S$ denotes the constant simplicial space $\underline p \mapsto S$, for some space $S$. This is easy to do. Next is where I get lost, (although I do understand that we're trying to write a general bisimplicial space $T$ is a direct limit of the special cases). He writes

"But any $T$ has a canonical presentation

$\coprod_{(r,s) \to (r',s')} h^{r',s'} > \times T_{rs} \rightrightarrows > \coprod_{(r,s)} h^{r,s} \times T_{rs} > \to T $

which is exact in the sense that the right arrow is the cokernel of the pair of arrows. Since the three functors from bisimplicial spaces to spaces under consideration commute with inductive limits, the lemma follows."

I am confused on a few points:

What is meant by 'canonical presentation'?

By 'cokernel', was 'coequaliser' meant?

In any case, how does one get an inductive limit from that diagram?

May someone please spell out what is going on here?

share|cite|improve this question
I should note that the paper of Tornehave that Quillen referenced in relation to the above seemingly never appeared. – Joshua Seaton Feb 20 '13 at 15:31
If my memory is correct, the result of Tornehave that Quillen referenced is part of Theorem A.4 in my "E_{\infty}$ spaces, group completions, and permutative categories.", #13 on my web page; the proof given there is ascribed to Tornehave's unpublished preprint. – Peter May Feb 20 '13 at 18:13
@Peter: I can't see any relationship between the above and your Theorem A.4, but that could very well be due to the fact that I am not yet versed in this stuff. – Joshua Seaton Feb 20 '13 at 19:54
I was remembering Quillen's reference, page 95, to May and Tornehave, which is in fact a reference to Theorem A.4. I didn't remember his earlier reference to Tornehave on page 94. – Peter May Feb 20 '13 at 20:09
Thank you anyway. – Joshua Seaton Feb 20 '13 at 21:23
up vote 3 down vote accepted

In the term "canonical presentation", "presentation" just means that the diagram you wrote is a coequalizer diagram presenting $T$ as being obtained from "simpler" bisimplicial spaces (so, yes, "cokernel of a pair" means "coequalizer"). This use of the term presentation comes from algebra, where presenting, say, a group, can be interpreted to mean writing it as a coequalizer of maps between free groups (and similarly for other algebraic structures).

The "canonical" part just means the whole diagram is a functor of $T$, or more concretely, that both $F(T):=\coprod_{(r,s) \to (r',s')} h^{r',s'} \times T_{rs}$ and $G(T) := \coprod_{(r,s)} h^{r,s} \times T_{rs}$ define (the object part) of functors and all three arrows in $F(T) \rightrightarrows G(T) \to T$ are components of natural transformations $F \rightrightarrows G \to \mathrm{Id}$.

Finally, the way the argument goes, once you have this functorial diagram, is that if $R$ denotes any of the three realization functors (from bisimplicial spaces all the way to spaces) you mention, then $R$ commutes with colimits (called "inductive limits" in the paper). This means that $R(F(T)) = \coprod_{(r,s) \to (r',s')} R(h^{r',s'} \times T_{rs})$, that $R(G(T)) = \coprod_{(r,s)} R(h^{r,s} \times T_{rs})$, and that the diagram

$$\coprod_{(r,s) \to (r',s')} R(h^{r',s'} \times T_{rs}) \rightrightarrows \coprod_{(r,s)} R(h^{r,s} \times T_{rs}) \to R(T)$$

is still a coequalizer diagram (but now in spaces). Since Quillen first proved that the spaces $R(h^{r,s} \times T_{pq})$ do not depend on which of the three $R$'s you take, the result, $R(T)$ does not depend on which $R$ you take either.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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