EDIT: removed cruft from this question.

Recall that extra degeneracies for an augmented simplicial set $X$ are maps $s_0\colon X_n \to X_{n+1}$ for $n=-1,0,1,2,\ldots$ which satisfy the usual simplicial identities with respect to the existing $d_i$, $s_i$. This definition clearly works for simplicial objects in pretty much an arbitrary category.

A simplicial set, augmented by $X_{-1} = \ast$, with extra degeneracies is contractible. In fact, given enough structure on an ambient category $C$, one can sensibly talk about homotopy of simplicial objects in $C$ (for example, one can say when $sC$ is a category with cofibrant objects, and has a notion of homotopy of maps). [EDIT: By this I mean there is some sort of model structure around relative to which we can talk about homotopy]

So my question is this: is it reasonable to think of simplicial objects in $C$ with extra degeneracies as being contractible for any category $C$ with terminal object? Certainly, ignoring size issues, we can think of such things as being contractible after we embed them in the category of simplicial sets in $Pre(C)$, if not some smaller (co)completion category.

Secondarily, can I get away with saying a simplicial object with extra degeneracies "is a contractible simplicial object?"


2 Answers 2


I don't have time to wade through terminology, but this seems to be an exercise in avoiding the use of simplicial homotopies, as defined for example in Definition 5.1 on page 12 of Simplicial Objects in Algebraic Topology. These homotopies make sense for simplicial objects in any ambient category $\mathcal{C}$ whatsoever, so that with this notion there is no such thing as a $\mathcal{C}$ "with insufficient structure to support homotopies''. Extra degeneracies give a particularly convenient way to construct just such a homotopy. If you have a terminal object, then you have a trivial map $\epsilon$ from any simplicial object $X$ to the constant simplicial object $\ast$ at the terminal object. A "base point'' is a map $\eta$ from $\ast$ to $X$ and is determined by a map in $\mathcal{C}$ from the terminal object to $X_0$. Then $\epsilon \circ \eta$ is the identity on $\ast$ and we say that $X$ is contractible if $\eta\circ \epsilon$ is homotopic to the identity. That makes sense in any $\mathcal{C}$, and it agrees with the usual notion in the usual examples.

  • $\begingroup$ My question is partly sociological, in that calling a simplicial object contractible when there is no model structure (of any flavour) is perhaps stretching the imagination. One can of course write down these definitions. If it is well-accepted such things can be called contractible, then I am satisfied. I can only guess that we think of them as being contractible as objects in some category with a class of weak equivalences. And you are right about all the stupid terminology - there was a second more complicated question in there that didn't quite make it out, and I should have cleaned up. $\endgroup$
    – David Roberts
    Jun 20, 2012 at 22:05

Actually the issue is not that simple. Here is a talk by Mike Barr on showing there are three different notions of 'contractible' for simplicial objects in a category that do not coincide even for simplicial sets. They even give explicit examples.

Michael Barr (joint with John F. Kennison, R. Raphael), Contractible simplicial objects, talk 9 October 2018 in the Logic and Categories seminar, McGill University, abstract, slides.

They find "Strong extra degeneracies" implies "Extra degeneracies" implies "Homotopic to a constant simplicial object", and these are strict implications. See the slides for precise definitions. Beware that the 'extra degeneracies' given in my question are not necessarily the same as Barr et al's definition.

  • 1
    $\begingroup$ There's also the possible notion that the simplicial object realizes to the terminal object (or the 0-th object if you want to consider the relative notion) in the nonabelian derived category (as defined, e.g., in Higher Topos Theory). This should be weaker than all of the ones mentioned in this question. $\endgroup$ Feb 5, 2019 at 13:19

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