Walk-through to Morel-Voevodsky A¹-homotopy theory part II (page 48, Lemma 1.1)

Wednesday, November 04th, 2009 | Author:

Here is the second part of my walk-through to Voevodskys A¹-homotopy theory:

On page 48, the first Lemma is shown. Without proof - so I will try to illuminate things a little bit by giving the proof. This lemma isn't used until section 3, so you can skip it, if you want to. I suggest not to do so, if you are intimidated by the diagram, because it isn't that hard, and it's a nice exercise to get the concepts in your head right.

A detailed explanation

The lemma states, that a monomorphism f : X \rightarrow Y between simplicial sheaves induces maps sk_{n-1}(f) \rightarrow sk_n(f) and Y_n \times \Delta^n \rightarrow sk_n(f) whose fiber coproduct ist the space

 ((X_n \coprod_{X_n^{deg}} Y_n^{deg}) \times \Delta^n) \coprod_{(X_n \coprod_{X_n^{deg}} Y_n^{deg})} (Y_n \times \partial\Delta^n)


What is \Delta^n?
First, there is a category \Delta, whose objects are just finite integer sequences \Delta_n := (0,1,...,n). One considers the opposed category \Delta^{op} (May writes \Delta^\ast instead), whose morphisms are defined to be the monotonic maps. Then \Delta := \Delta^{op\ op}. It is important to understand the nature of these monotonic maps - read at least the first chapter in Mays "Simplicial Objects in Algebraic Topology". You will then see, that a simplicial set ist just a functor \Delta^{op} \rightarrow Sets, just as any simplicial object ist a functor from \Delta^{op}.
For each object \Delta_n \in \Delta, define \Delta^n := Hom(-,\Delta_n), a functor \Delta^{op} \rightarrow Sets, thus a simplicial set. Together with the embedding of sets as constant sheafs, this gives us a simplicial sheaf \Delta^n, which is what is meant here. Observe that we have a functor \Delta \rightarrow \Delta^{op}Sh(T) by sending a map \alpha : \Delta_n \rightarrow \Delta_m to the morphism \alpha \circ : \Delta^n \rightarrow \Delta^m. This functor \Delta^\bullet is a cosimplicial object and turns the category of simplicial sheaves on the site T into a simplicial model category (that is, a model category which is compatibly enriched over simplicial sets, so the Hom-Sets are simplicial sets) with the simplicial function object

 S(X,Y) = Hom_{\Delta^{op}Sh(T)}(X \times \Delta^\bullet, Y)

We will talk in detail about the simplicial model category structure later.

So what is sk_n(f)?
It's defined to be sk_n(f) := f(X) \cup sk_n(Y) where sk_{-1}(Y)=\emptyset and for n \geq 0 is sk_n(Y) the n-skeleton of Y, that is the image of the "obvious" functor X_n \times \Delta^n \to X, which may not be so obvious at first sight. Remember that X and \Delta^n are both contravariant functors into the category of simplicial sets, so by Yoneda lemma we have Trans(\Delta^n, X) = X(\Delta^n) = X_n. This gives us an interpretation of X_n \times \Delta^n as Hom(\Delta^n, X) \times \Delta^n, where the Hom is taken in the category of functors. The "obvious" functor X_n \times \Delta^n \to X is the evaluation map. Observe that (by Yoneda Lemma) for x \in X_n we have \overline{x}(\Delta_n) = x, where \overline{x} \in Trans(\Delta^n, X) corresponds to x.
At the moment, we don't care about the right adjoint cosk_n (coskeleton) of the skeleton functor.

What is X_n^{deg}?
It's defined to be X_n^{deg} := \cup_{i=0}^{n-1} s_i^{n-1}(X_{n-1}), the union of all degenerated n-simplices that come from X_{n-1} (therefore the name deg). In Mays book "Simplicial Objects", this is written \bar X_n. Another equivalent definition would be X_n^{deg} = (sk_{n-1}(X))_n.

The statement of the lemma explained

What are these complicated looking objects in the diagram?
Let's look at the following pushout diagram:
pushout of f : X_n^deg -> Y_n^deg and X_n^deg -> X_n, denoted by Z_n.
Where I have given the name Z_n := X_n \coprod_{X_n^{deg}} Y_n to make it look less scary. The letter \iota denotes inclusion. There is clearly an inclusion Y_n^{deg} \rightarrow Y_n and the restriction of f gives an arrow X_n \rightarrow Y_n and by pushout property we get a unique arrow \phi : Z_n \to Y_n that makes the diagram commutative:
pushout property gives arrow phi : Z_n -> Y_n
The next step is again a pushout diagram:
pushout of phi x id : Z_n x del Delta^n --> Y_n x del Delta^n and id x iota : Z_n x del Delta^n --> Z_n x Delta^n, denoted by W_n
Where I have given the name W_n := (Z_n \times \Delta^n) \coprod_{(Z_n \times \partial \Delta^n)} (Y_n \times \partial \Delta^n). We have morphisms id \times \iota : Y_n \times \partial \Delta^n \rightarrow Y_n \times \Delta^n and \phi \times id : Z_n \times \Delta^n \to Y_n \times \Delta^n that give us by the pushout property a unique arrow \psi : W_n \rightarrow Y_n \times \Delta^n.
pushout property gives arrow psi : W_n -> Y_n x Delta^n

What are the maps in the cocartesian diagram?
The arrow \psi constructed above is the left arrow of the diagram. The right arrow can only be an inclusion sk_{n-1}(f) \rightarrow sk_n(f). The bottom arrow is just the evaluation morphism explained above ev : Y_n \times \Delta^n \rightarrow sk_n(Y) \subseteq sk_n(f).
To construct the top arrow, look at the pushout defining W_n above. The composition of \phi \times id : Z_n \times \Delta^n \rightarrow (f(X_n) \cup Y_n^{deg}) \times \Delta^n with the evaluation morphism Y_n \times \Delta^n \rightarrow sk_n(Y) and the evaluation morphism Y_n \times \partial \Delta^n \rightarrow sk_{n-1}(Y) give us, by pushout property, a unique morphism \alpha : W_n \rightarrow sk_{n-1}(f).


To check that the diagram is cocartesian means to check that sk_{n}(f) has the pushout property in the diagram
simplified diagram of Lemma 1.1, pushout of alpha : W_n -> sk_{n-1}(f) and psi : W_n -> Y_n x \Delta^n
Take a test object V together with morphisms s : sk_{n-1}(f) \rightarrow V and t : Y_n \times \Delta^n \rightarrow V such that s\circ \alpha = t \circ \psi. We have to find a unique morphism \beta : sk_n(f)\rightarrow V that makes the whole diagram commutative.

For a moment, think about the sheaves as sheaves on T = pt, so our category is just the category of simplicial sets. We can prove the theorem in this case:

Remember sk_n(f) = f(X) \cup sk_n(Y). For each \sigma \in sk_{n-1}(Y) we can define \beta(\sigma) := s(\sigma). For \sigma \in sk_n(Y) we see that \sigma lies in the image of the (monic) evaluation map from Y_n \times \Delta^n, so we can define \beta(\sigma) := t(ev^{-1}(\sigma)). Is \beta well-defined? If \sigma \in sk_n(f) has a preimage under the evaluation map, it lies in sk_n(Y); if it has at the same time a preimage of sk_{n-1}(f), it lies in sk_{n-1}(Y), which lies in the image of the map \alpha, therefore we seee that \beta is well-defined.

Uniqueness? If we'd have a second map \beta' : sk_n(f) \to V, but \beta \neq \beta', then we would have a \sigma \in sk_n(f) such that \beta(\sigma) \neq \beta'(\sigma). Such a \sigma can't have a preimage under the inclusion of s_{n-1}(f) or the evaluation on Y_n \times \Delta^n. So \sigma \notin f(X) and \sigma \notin sk_n(Y). Nothing remains, so \sigma doesn't exist.

The general case is done by applying all points x^\ast : Sh(T) \rightarrow Sets respectively all points x^\ast : \Delta^{op}Sh(T) \rightarrow \Delta^{op}Sets, which are finite limit- and finite colimit-preserving functors. We can apply the lemma for the point case (which we have proved just before) and deduce from its validity for all points its validity in general (since our site T has enough points - we took only T like that in the beginning).


5 Responses

  1. It's great to see somebody blogging about this stuff! Just a minor comment on the last paragraph: in general, points will only commute to _finite_ limits.
    I'm looking forward to part III :-)

  2. Thanks for the correction of the last paragaph. My language is sometimes lazy when it comes to smallness & finiteness issues in categories. It's good to have someone proof-reading :-)

    I'm working on part III, but sadly there is a lot of other stuff to do...

  3. I didn't write part III, but now there is a post about some categorical background needed to understand properly what's going on:


  4. 4
    Rakesh Pawar 
    2016-10-05 (5. October 2016)

    You take \Delta^n as the constant sheaf "associated" to the simplicial set \Delta^n.
    On the other hand, after following the definition of product of a simplicial object and a simplicial set as in http://stacks.math.columbia.edu/download/simplicial.pdf
    Definition 13.1, I was wondering if we should think about X \times \Delta^n as the product via this definition, which is a simplicial sheaf as X is so.
    At this point I am just puzzled at the correct meaning of X \times \Delta^n. Your definition surely makes sense (via constant sheaf). But what about the definition from stacks project?
    Anyways thanks for the post.

  5. the correct meaning of X x Y is the categorical product, which (if it exists) is unique. The definition from stacks project satisfies the universal property.

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