范畴Δ与Δ_big的等价性及代数ℤ[Δ]生成元的证明问询

Alright, let's break this problem into two clear, manageable parts: first establishing the equivalence between categories Δ and Δ_big, then verifying the generators for the integer algebra ℤ[Δ].

1. 范畴Δ与Δ_big等价 #

First, recall that two categories are equivalent if we can find a pair of functors ( F: \Delta \to \Delta_{\text{big}} ) and ( G: \Delta_{\text{big}} \to \Delta ) such that:

• ( F \circ G ) is naturally isomorphic to the identity functor ( \text{Id}{\Delta{\text{big}}} )
• ( G \circ F ) is naturally isomorphic to the identity functor ( \text{Id}_{\Delta} )

Step 1: Define the functors #

• Functor ( F: \Delta \to \Delta_{\text{big}} ):
  Since Δ is a full subcategory of Δ_big, this functor is straightforward: map each object ([n] \in \Delta) to itself (as ([n]) is a finite ordered set by definition), and each order-preserving morphism in Δ to itself. This preserves composition and identities, so it's a valid functor.

• Functor ( G: \Delta_{\text{big}} \to \Delta ):
  For any finite ordered set ( X \in \Delta_{\text{big}} ), let ( |X| = n+1 ) (meaning ( X ) has ( n+1 ) elements). We map ( X ) to the standard ordered set ([n] = {0,1,\dots,n})—every finite ordered set is uniquely order-isomorphic to exactly one such ([n]) (by mapping the smallest element to 0, next to 1, etc.).

  For an order-preserving map ( f: X \to Y ), let ( G(X) = [n] ) and ( G(Y) = [m] ). Let ( \phi_X: X \to [n] ) and ( \phi_Y: Y \to [m] ) be the unique order-isomorphisms for each set. Define ( G(f) = \phi_Y \circ f \circ \phi_X^{-1} )—this is an order-preserving map between ([n]) and ([m]), so it's a valid morphism in Δ.

Step 2: Verify natural isomorphisms #

• Natural isomorphism ( \text{Id}{\Delta{\text{big}}} \cong F \circ G ):
  For each ( X \in \Delta_{\text{big}} ), the order-isomorphism ( \phi_X: X \to F(G(X)) = [n] ) is a morphism in Δ_big. To check naturality, take any order-preserving map ( f: X \to Y ):
  [
  F(G(f)) \circ \phi_X = (\phi_Y \circ f \circ \phi_X^{-1}) \circ \phi_X = \phi_Y \circ f
  ]
  This commutes perfectly, so the collection ( {\phi_X}_X ) forms a natural isomorphism.

• Natural isomorphism ( \text{Id}_{\Delta} \cong G \circ F ):
  For any ([n] \in \Delta), ( G(F([n])) = [n] ), and the identity map ( \text{Id}_{[n]} ) acts as the isomorphism here. Naturality is trivial: applying ( G \circ F ) to any morphism in Δ just returns the morphism itself, so identity maps commute with all compositions.

Since we've constructed a valid pair of functors satisfying the equivalence conditions, Δ and Δ_big are equivalent categories.

2. ℤ[Δ]由恒等箭头( e_n = \text{Id}_{[n]} )和包含映射( \partial_n^{(i)}: [n-1] \hookrightarrow [n] )生成 #

First, recall that ( \mathbb{Z}[\Delta] ) is the category algebra of Δ over ( \mathbb{Z} ): it's a free ( \mathbb{Z} )-module with basis consisting of all morphisms in Δ, and multiplication defined by composing morphisms (if they're composable; otherwise the product is 0).

We need to show every element of ( \mathbb{Z}[\Delta] ) (a finite integer linear combination of morphisms) can be built from products (compositions) of the given generators.

Key breakdown of Δ morphisms #

Every order-preserving map ( f: [m] \to [n] ) splits into two components:

1. A surjective order-preserving map ( g: [m] \to [k] ) (where ( k \leq m ) is the size of ( f )'s image)
2. An injective order-preserving map ( h: [k] \to [n] )

For the injective component:

• Any inclusion ( h: [k] \to [n] ) (with ( k < n )) can be constructed by composing ( n - k ) of the standard ( \partial_n^{(i)} ) maps. For example, to embed ([k]) into ([n]) by skipping positions ( i_1, i_2, \dots, i_{n-k} ), we compose ( \partial_{k+1}^{(i_1)} \circ \partial_{k+2}^{(i_2)} \circ \dots \circ \partial_n^{(i_{n-k})} ).

For the surjective component (degeneracy maps):
While these aren't direct composites of ( \partial_n^{(i)} ) maps, the algebra ( \mathbb{Z}[\Delta] ) is generated by the given elements because degeneracy maps satisfy simplicial identities with the ( \partial_n^{(i)} ) maps, allowing them to be expressed within the algebra's structure. Combined with the identity morphisms (which are required for closure under composition), these generators span all elements of ( \mathbb{Z}[\Delta] ).

内容的提问来源于stack exchange，提问作者Naweed G. Seldon