Recall Waldhausen's fibration theorem, called 'localization theorem' in Thomason-Trobaugh: if $v\subset w\subset \mathcal C$ are categories of weak equivalences in a category with cofibrations $\mathcal C$, then under some hypotheses of saturation, extension and cylinder, the inclusions yield a homotopy fibration sequence:

$Kv\mathcal C^w\to Kv\mathcal C\to Kw\mathcal C$.

The specified nullhomotopy is given as follows: there is a weak equivalence from 0 to the inclusion $\mathcal vC^w\to w\mathcal C$ that thus induces a homotopy on $K$-theory (prop. 1.3.1 Waldhausen).

My question is the following. Suppose we have another $v'\subset w'\subset \mathcal C'$ under the same hypotheses, and suppose we have an exact functor $F:\mathcal C\to \mathcal C'$ sending $v$ to $v'$ and $w$ to $w'$. Then consider the two homotopy fibration sequences given by the theorem above; they fit into a strictly commutative ladder diagram.

Now take the homotopy fiber of the second map in both homotopy fibrations. Call them $X$ and $X'$. Considering a functorial construction of homotopy fibers, we have another strictly commutative ladder diagram.

Consider the induced maps $Kv\mathcal C^w\to X$ and $Kv'\mathcal C'^{w'}\to X'$. Do they make the following square strictly commute?

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  • $\begingroup$ If I am not clear enough let me know and I'll draw another diagram. $\endgroup$ Commented May 7, 2014 at 15:18

1 Answer 1


Basically yes. Introduce the contractible $Kw\mathcal C^w$ into the picture, so that $Kv\mathcal C^w$ gets replaced by the equivalent homotopy fiber of $Kv\mathcal C^w\to Kw\mathcal C^w$.


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