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Waldhausen's definition of a category with cofibrations includes the choice of a distinguished zero object. Probably this also means that an exact functor should preserve zero objects on the nose (Waldhausen writes "takes $\*$ to $\*$"). Isn't this rather evil from the general perspective of category theory? Even if we had fixed zero objects, we should only require that an exact functor preserves them up to isomorphism (which is unique anyway).

Remark that Weibel's K-Book doesn't demand this choice; there it is only required that some zero object exists. Isn't this more natural? On the other hand, the theories look quite the same.

So a question might be why Waldhausen has chosen his definition and if one definition of the two definitions has any real advantage in practice against the other.

This question has a topological analogue. If $(X,x)$ and $(Y,y)$ are pointed spaces, then why not defining a map $(X,x) \to (Y,y)$ to be a map $f : X \to Y$ together with a path $f(x) \to y$ (oplax) or $y \to f(x)$ (lax)? This has the advantage that we are more flexible as concerning the base points, we still have a well-defined induced map $\pi_n(X,x) \to \pi_n(Y,y)$ on homotopy groups, but on the other hand this will be more complicated than the usual notion of a pointed map. Any references are welcome.

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    $\begingroup$ Regarding your question about topological spaces: If you take this definition, then you don't have a composition of two maps which is associative. Instead, you either need to switch to a less straightforward "Moore path" style definition or accept that the object is instead an $A_\infty$ category. If we're generalizing that far, why don't we replace points with contractible spaces? For many applications, the responsibility for maintaining the definition outweighs the added utility or increase in aesthetic value. $\endgroup$ Apr 9, 2012 at 13:09
  • $\begingroup$ Thanks for this comment. This also reflects the situation with pointed categories: They form a $2$-category. $\endgroup$ Apr 9, 2012 at 13:24
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    $\begingroup$ You seem to be asking about people's opinion. Wouldn't you like to make a precise question? $\endgroup$ Apr 9, 2012 at 18:40
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    $\begingroup$ @Downvoters: Please explain how I can improve this question. $\endgroup$ Apr 10, 2012 at 7:06
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    $\begingroup$ Martin - I didn't vote your question down, but I don't see why those who did need to explain their opinion, any more than up-voters need to explain theirs. I suspect that John Klein's answer gives some idea of the down-voters motivations, anyway. $\endgroup$
    – HJRW
    Apr 10, 2012 at 9:35

2 Answers 2

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When Waldhausen defines $K$-theory, he considers sequences of the form

$* \rightarrow A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n$

for varying $n$, with all maps cofibrations. For any fixed $n$, these sequences together with the obvious morphisms again form a category, and these categories nearly assemble to a simplicial object in categories. Degeneracy maps are defined by inserting identities and the face maps except $d_0$ are defined by composition. $d_0$ of the above sequence is obtained by ignoring the $*$ and dividing out $A_1$ from the sequence to obtain

$A_1/A_1 \rightarrow A_2/A_1 \rightarrow \cdots $

However, this does not really make sense since choices of quotients are involved; in particular a choice of zero object $A_1/A_1$. You may make arbitrary choices, but then you only obtain a simplicial object up to equivalence: For example, $d_0 s_0$ is not the identity functor, but merely isomorphic to it.

This problem is solved by considering a different category: We plug more information into our objects by declaring an object to be a sequence

$* \rightarrow A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n$

together with choices of quotients $A_i/A_j$, $i \geq j$. This yields an equivalent category, but now we have a shot to get an honest simplicial object in categories.

But then a problem in the degeneracy maps arises: In a degenerate sequence of the form

$* \rightarrow \cdots A_k = A_k \rightarrow \cdots$

we now also have to specify the quotient $A_k/A_k$, and again making arbitrary choices will not cut it. Letting this quotient be the chosen zero object works when one also restricts to sequences starting with this chosen zero object. I think this is the main technical reason to pick a zero object once and for all.

Such evil concepts are actually quite common in $K$-theory: Often, it is not enough to have diagrams of categories commuting up to natural isomorphism, so one has to wriggle the definitions of the involved categories a bit to obtain different, equivalent categories with a strictly commuting diagram of functors.

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  • $\begingroup$ @Fabi: Thanks a lot! This answers my question. I will accept your answer in the next days. $\endgroup$ Apr 10, 2012 at 7:10
  • $\begingroup$ If one triesdto be non-evil at any cost, one could ask the following: Is it not more sensible to replace the category of categories (where one has this simplicial object) by the infinity-category of categories? Probably it gives a simplicial object there, well-defined up to contractible choice. $\endgroup$ Apr 10, 2012 at 8:35
  • $\begingroup$ @Lennart Analogous to how a stable (oo,1)-category is what a triangulated category is trying to be, and solves the functoriality issues? $\endgroup$
    – David Roberts
    Apr 10, 2012 at 9:26
  • $\begingroup$ How do you make a space/simplicial set/bisimplicial set out of a "infinity-simplicial" object in categories? $\endgroup$ Apr 10, 2012 at 10:41
  • $\begingroup$ You can always strictify an "infinity-diagram" in an infinity-category coming from a model category to a strict diagram in the model category, which is well-defined up to contractible choice. If my memory doesn't fail me, one can also directly geometrically realize a infinity-simplicial object, say, in the infinity-category of simplicial sets to a simplicial set (well defined up to contractible choice). Note that I don't claim that it is usually sensible to take this route. $\endgroup$ Apr 16, 2012 at 13:48
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Although it might seem surprising to some readers, Waldhausen was not going for abstraction merely for the sake of itself in making his definitions: he had concrete applications in mind (most importantly, to understanding manifolds). It was the applications that drove the definitions, not the other way around.

Remark

You can avoid choosing quotient data (as well as a specific zero object) if you instead use Thomason's $wT_\bullet$-construction.

In the latter simplicial category, objects in degree $n$ are strings of cofibrations $$ A_\bullet = A_0 \to A_1 \to \cdots\to A_n $$ (here $A_0$ needn't be the zero object) and a morphism is a map of strings $A_\bullet \to B_\bullet$ in which each square $$ A_k \to A_{k+1} $$ $$ \downarrow \qquad\downarrow $$ $$ B_k \to B_{k+1} $$ is a pushout up to weak equivalence meaning that the map $$ B_k \cup_{A_k} A_{k+1} \to B_{k+1} $$ is a weak equivalence, where the domain of the latter denotes any (non-specified) choice of pushout for $B_k \leftarrow A_k \to A_{k+1}$.

(Note: it is not necessarily the case that the maps $A_k \to B_k$ are weak equivalences, but if $A_0 \to B_0$ is a weak equivalence, then the axioms imply that $A_k \to B_k$ is too for each $k$.)

The face operators are given by removing a term in the filtration and the degeneracies are given by inserting the identity. This gives a simplicial category which has the same homotopy type as Waldhausen's $wS_\bullet$-construction, but where we haven't used the zero object nor any quotient data.

One can find this construction in Waldhausen's foundational paper (linked to above). The construction is important, for example, when one uses the "manifold approach" to defining $A(X)$.

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    $\begingroup$ Or in an analogy: when you hike through the wilderness you don't build a paved, divided highway on your very first trek. You build the minimum trail you need to get to your destination. As population density increases, people enlarge the trail to a road, build power lines, ditches, surface the road, cover it with gravel, eventually one might pave it and maybe years later it becomes a highway. $\endgroup$ Apr 10, 2012 at 0:39
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    $\begingroup$ Of course I understand what you are saying. But this does not really answer my question and does not address Waldhausen categories specifically. What is the precise motivation for this choosen zero object? $\endgroup$ Apr 10, 2012 at 6:49
  • $\begingroup$ +10000 ${}{}{}{}$ $\endgroup$ Apr 10, 2012 at 6:53
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    $\begingroup$ By the way, I don't like at all the tone of this answer and its comments which suggest that I have misunderstood Waldhausen's motivations, without giving any specific information. $\endgroup$ Apr 10, 2012 at 7:16
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    $\begingroup$ Martin: I tried to answer your question! You asked "Why had chosen his definition..." I was trying to point out that the definition was not prompted by abstract considerations. I was not trying to offend. $\endgroup$
    – John Klein
    Apr 10, 2012 at 11:18

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