03.12.E5 · modern-geometry / homotopy

Simplicial objects and Dold-Kan exercise pack (May supplement)

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Anchor (Master):

Formal definition of the pack Intermediate

May's Simplicial Objects in Algebraic Topology develops the algebra of functors $Δ^{op} \to A$ into a target category $A$ . When $A = Ab$ , the resulting simplicial abelian groups are equivalent to non-negatively graded chain complexes via the Dold-Kan correspondence — the structural theorem that homological algebra and simplicial algebra are the same subject in degree $\geq 0$ . The book also builds the bar construction, which turns a group (or monoid, or algebra) into a simplicial object whose realization is a classifying space, and proves the Eilenberg-Zilber theorem comparing the chains on a product with the tensor product of chains.

This pack collects nine exercises drawn from §17-§31 — two easy, four medium, three hard — each with a hint and a full solution. The exercises are grouped: the normalized/Moore complex and degeneracies (easy), the Dold-Kan functors and the bar construction (medium), and the Eilenberg-Zilber theorem with its shuffle and Alexander-Whitney maps, plus the homotopy of a simplicial abelian group (hard).

The conventions are May's: for a simplicial abelian group $A$ , the Moore complex $(C_{*} A, \partial)$ has $C_{n} A = A_{n}$ with $\partial = \sum_{i} (- 1)^{i} d_{i}$ ; the normalized complex $N_{n} A = ⋂_{i > 0} ker d_{i}$ with differential $d_{0}$ ; and the degenerate subcomplex $D_{n} A$ is generated by the images of the degeneracies $s_{i}$ . Homotopy of a simplicial abelian group means homology of its associated chain complex: $π_{n} A = H_{n} (C_{*} A)$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as the model solution. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. State the Dold-Kan correspondence and prove the normalization theorem $C_ A = N_* A \oplus D_* A $w i t h$ D_* A$ acyclic.*

Solution. Dold-Kan correspondence. The normalized-chains functor $N : sAb \to Ch_{\geq 0} (Ab)$ and the functor $Γ : Ch_{\geq 0} \to sAb$ are mutually inverse equivalences of categories. Under it, $π_{n} A = H_{n} (N A)$ , and the equivalence is natural.

Normalization theorem. For a simplicial abelian group $A$ , the Moore complex splits as $C_{*} A = N_{*} A \oplus D_{*} A$ , where $N_{n} A = ⋂_{i \geq 1} ker d_{i}$ is the normalized subcomplex and $D_{n} A = \sum_{i} im s_{i}$ is the degenerate subcomplex, and the inclusion $N_{*} A ↪ C_{*} A$ is a chain-homotopy equivalence because $D_{*} A$ is acyclic.

Proof. First, $C_{n} A = N_{n} A \oplus D_{n} A$ as abelian groups: the simplicial identities let one write any element uniquely as a sum of a normalized element and a sum of degenerate ones, by an explicit idempotent built from the $s_{i} d_{i}$ . Concretely, the operators $P^{(k)} = \prod (1 - s_{i} d_{i})$ assembled in order give a projection of $C_{n} A$ onto $N_{n} A$ along $D_{n} A$ . The differential $\partial = \sum_{i} (- 1)^{i} d_{i}$ restricts to $N_{*} A$ (since on the normalized part all $d_{i}$ with $i \geq 1$ vanish, leaving $d_{0}$ up to sign) and preserves $D_{*} A$ .

The degenerate complex $D_{*} A$ is acyclic: an explicit contracting homotopy is built from the extra degeneracy / the simplicial identities $d_{i} s_{i} = id$ , which shows every cycle in $D_{*}$ is a boundary. Hence the projection $C_{*} A \to N_{*} A$ is a quasi-isomorphism — in fact a chain-homotopy equivalence — and $H_{*} (C_{*} A) = H_{*} (N_{*} A)$ . $□$

This is the engine of the whole subject: it lets one compute the homotopy of a simplicial abelian group with the small normalized complex, and it is the technical core of the Dold-Kan equivalence.

Exercises Intermediate

Exercise 1 (easy, proof). Simplicial identities for faces and degeneracies.

Write down the simplicial identities relating the face maps $d_{i}$ and degeneracy maps $s_{j}$ , and verify the identity $d_{i} s_{i} = id = d_{i + 1} s_{i}$ .

Hint

There are five families. The ones you need say "a degeneracy followed by its adjacent face is the identity."

Answer

The simplicial identities are:

d_{i} d_{j} s_{i} s_{j} d_{i} s_{j} d_{i} s_{j} d_{i} s_{j} = d_{j - 1} d_{i} = s_{j + 1} s_{i} = s_{j - 1} d_{i} = id = s_{j} d_{i - 1} (i < j), (i \leq j), (i < j), (i = j or i = j + 1), (i > j + 1) .

The fourth family gives directly $d_{j} s_{j} = id$ and $d_{j + 1} s_{j} = id$ ; setting $i = j$ in the first case and $i = j + 1$ in the second. Geometrically, $s_{j}$ inserts a degenerate (repeated) vertex, and the two adjacent faces $d_{j}, d_{j + 1}$ delete that repeated vertex, returning the original simplex. These two identities are what make the degenerate subcomplex acyclic and power the normalization theorem. May §1, §17.

Exercise 2 (easy, numeric). Dold-Kan on a chain complex concentrated in one degree.

Apply the Dold-Kan functor $Γ$ to the chain complex with $A$ in degree $n$ and $0$ elsewhere. Identify the resulting simplicial abelian group.

Hint

$Γ$ of $A [n]$ is the simplicial abelian group $K (A, n)$ — the Eilenberg-MacLane object.

Answer

The chain complex $A [n]$ (the group $A$ in degree $n$ , zero elsewhere) corresponds under $Γ$ to the Eilenberg-MacLane simplicial abelian group $K (A, n)$ .

By the Dold-Kan dictionary, its homotopy groups equal the homology of the chain complex:

π_{k} (Γ A [n]) = H_{k} (A [n]) = {A 0 k = n k \neq = n .

So $Γ A [n] = K (A, n)$ , the simplicial model of the Eilenberg-MacLane space whose realization is the topological $K (A, n)$ . Explicitly, $Γ A [n]$ in simplicial degree $m$ is a sum of copies of $A$ indexed by the surjections $[m] ↠ [n]$ in $Δ$ — the degeneracy operators of the simplex category. This is the canonical example showing Dold-Kan converts the one-term chain complex into the building block of all homology. May §22, §23.

Exercise 3 (medium, proof). $\pi_$ of a simplicial abelian group is its normalized homology.*

Prove that for a simplicial abelian group $A$ , the homotopy groups $π_{n} (A)$ (defined simplicially via the Kan condition) coincide with the homology of the normalized chain complex $H_{n} (N A)$ .

Hint

Every simplicial abelian group is a Kan complex. Identify the cycles and boundaries of $N A$ with simplicial homotopy classes.

Answer

Step 1: $A$ is a Kan complex. A simplicial abelian group satisfies the Kan condition: any horn $Λ_{k}^{n} \to A$ extends, because one can solve the horn-filling equations linearly using the group structure and the simplicial identities (Moore's theorem). So the simplicial homotopy groups $π_{n} (A, 0)$ are defined.

Step 2: identify cycles. An element of $π_{n} (A)$ is represented by an $n$ -simplex $α$ with all faces zero, i.e. $d_{i} α = 0$ for all $i$ . Such $α$ lies in $N_{n} A = ⋂_{i \geq 1} ker d_{i}$ and additionally has $d_{0} α = 0$ — exactly a cycle of the normalized complex (whose differential is $d_{0}$ on $N_{*}$ ).

Step 3: identify the relation. Two such representatives $α, α^{'}$ are simplicially homotopic iff their difference $α - α^{'}$ is $d_{0}$ of a normalized $(n + 1)$ -chain — exactly a boundary in $N_{*} A$ . The homotopy is built from an $(n + 1)$ -simplex whose faces encode the homotopy, which the linearity again lets one normalize.

Hence $π_{n} (A) = ker (d_{0} : N_{n} A \to N_{n - 1} A) / im (d_{0} : N_{n + 1} A \to N_{n} A) = H_{n} (N A)$ . The group structure on $π_{n}$ matches the additive structure on homology. This is the bridge theorem: simplicial homotopy of an abelian object = chain homology. May §17, §22.

Exercise 4 (medium, proof). The bar construction of a group.

Describe the bar construction $B_{∙} G$ of a (discrete) group $G$ as a simplicial set, and show its realization $B G = ∣ B_{∙} G ∣$ is a $K (G, 1)$ .

Hint

$B_{n} G = G^{n}$ , with face maps multiplying or dropping adjacent entries. This is the nerve of $G$ as a one-object groupoid.

Answer

The bar construction $B_{∙} G$ has $n$ -simplices $B_{n} G = G^{n} = {[g_{1} ∣ g_{2} ∣ \dots ∣ g_{n}]}$ (tuples of group elements, "bars" separating them). The face maps are

d_{i} [g_{1} ∣ \dots ∣ g_{n}] = ⎩ ⎨ ⎧ [g_{2} ∣ \dots ∣ g_{n}] [g_{1} ∣ \dots ∣ g_{i} g_{i + 1} ∣ \dots ∣ g_{n}] [g_{1} ∣ \dots ∣ g_{n - 1}] i = 0 0 < i < n i = n,

and degeneracies insert an identity element. This is exactly the nerve $N G$ of $G$ regarded as a one-object groupoid.

$B G$ is a $K (G, 1)$ . Since $G$ is a groupoid (every element invertible), $N G = B_{∙} G$ is a Kan complex. Its realization is connected with $π_{1} (B G) = G$ (loops are words in the generators, relations are the $2$ -simplices recording multiplication). The higher homotopy vanishes: the universal cover of $B G$ is the realization of the contractible simplicial set $E_{∙} G$ (with $E_{n} G = G^{n + 1}$ ), so $π_{n} (B G) = 0$ for $n \geq 2$ . Hence $B G ≃ K (G, 1)$ , the classifying space.

The bar construction generalizes to a two-sided bar $B (M, G, N)$ for a monoid $G$ acting on $M, N$ , the universal device for homotopy quotients and classifying spaces of group actions. May §29-§31.

Exercise 5 (medium, proof). Homology of the bar construction is group homology.

Show that the homology of the chain complex of the bar construction $B_{∙} G$ computes the group homology $H_{*} (G; Z)$ .

Hint

The normalized chains on $B_{∙} G$ form the standard bar resolution computing $Tor_{*}^{Z G} (Z, Z)$ .

Answer

Apply the free-abelian-group functor $Z [-]$ to the bar construction to get a simplicial abelian group $Z [B_{∙} G]$ , then take its associated (normalized) chain complex $C_{*}$ . By definition $H_{*} (B G; Z) = H_{*} (C_{*})$ .

The normalized complex of $Z [B_{∙} G]$ is precisely the standard (inhomogeneous) bar resolution: $C_{n} = Z [G^{n}]$ with differential $\partial = \sum_{i} (- 1)^{i} d_{i}$ , where the $d_{i}$ are the bar face maps of Exercise 4. This is the canonical free resolution of $Z$ as a $Z G$ -module with the identity $G$ -action, after tensoring down with $Z$ over $Z G$ . Therefore

H_{n} (B G; Z) = H_{n} (C_{*}) = Tor_{n}^{Z G} (Z, Z) = H_{n} (G; Z),

the group homology of $G$ . Dually, cohomology of $B G$ is group cohomology $H^{*} (G; Z)$ .

This is the topological incarnation of group (co)homology: the classifying space $B G = K (G, 1)$ has (co)homology equal to the algebraically defined group (co)homology, and the bar construction is the explicit chain-level bridge. May §29; Weibel §8.4.

Exercise 6 (medium, numeric). Shuffle map on low-degree simplices.

The Eilenberg-Zilber shuffle map $\nabla : C_{p} (X) \otimes C_{q} (Y) \to C_{p + q} (X \times Y)$ sends a tensor of simplices to a signed sum of $(p, q)$ -shuffles. Write $\nabla$ explicitly for $p = q = 1$ , the product of two $1$ -simplices.

Hint

A $(1, 1)$ -shuffle of a square is its two triangles. The two $(1, 1)$ -shuffles of $[01]$ are $(0) (1)$ orderings of the two degeneracy directions.

Answer

For $p = q = 1$ , the shuffle map sends $x \otimes y$ (with $x \in C_{1} X$ , $y \in C_{1} Y$ ) to a sum over the two $(1, 1)$ -shuffles, which triangulate the square $Δ^{1} \times Δ^{1}$ into two $2$ -simplices:

\nabla (x \otimes y) = (s_{1} x \times s_{0} y) - (s_{0} x \times s_{1} y),

where $s_{0}, s_{1}$ are the degeneracies producing the two prisms over the edges, and the sign is the shuffle sign $sgn (μ)$ . Geometrically: the square with corners $(x_{0}, y_{0}), (x_{1}, y_{0}), (x_{0}, y_{1}), (x_{1}, y_{1})$ splits along a diagonal into two triangles, and $\nabla$ records this triangulation with the standard orientation signs.

In general $\nabla (x \otimes y) = \sum_{(μ, ν)} sgn (μ, ν) (s_{ν} x \times s_{μ} y)$ summed over $(p, q)$ -shuffles $(μ, ν)$ of ${0, \dots, p + q - 1}$ . The shuffle map is a chain map and, paired with the Alexander-Whitney map, gives the Eilenberg-Zilber equivalence. May §29.

Exercise 7 (hard, proof). The Eilenberg-Zilber theorem.

State the Eilenberg-Zilber theorem and prove it via acyclic models, identifying the two natural chain-homotopy-inverse maps.

Hint

The shuffle map $\nabla$ and the Alexander-Whitney map $A W$ are chain-homotopy inverse. Acyclic models supplies the homotopies because both sides agree on representables and are acyclic in positive degrees.

Answer

Theorem (Eilenberg-Zilber). For simplicial sets (or spaces) $X, Y$ , the chain complexes $C_{*} (X \times Y)$ and $C_{*} (X) \otimes C_{*} (Y)$ are naturally chain-homotopy equivalent, via the Alexander-Whitney map $A W : C_{*} (X \times Y) \to C_{*} (X) \otimes C_{*} (Y)$ and the shuffle map $\nabla : C_{*} (X) \otimes C_{*} (Y) \to C_{*} (X \times Y)$ , which are mutually chain-homotopy inverse: $A W \circ \nabla = id$ and $\nabla \circ A W ≃ id$ .

The Alexander-Whitney map is the "front face $\otimes$ back face" formula

A W (σ) = p + q = n \sum (d_{p + 1} \dots d_{n} σ) \otimes (d_{0}^{p} σ),

and $\nabla$ is the shuffle map (Exercise 6).

Proof by acyclic models. Both functors $F = C_{*} (- \times -)$ and $G = C_{*} (-) \otimes C_{*} (-)$ from $sSet \times sSet$ to chain complexes are free on the models ${(Δ^{p}, Δ^{q})}$ (their chains are generated by the images of the universal simplices) and acyclic on models (the chains of $Δ^{p} \times Δ^{q}$ and of $C_{*} Δ^{p} \otimes C_{*} Δ^{q}$ are both contractible above degree $0$ , since products and tensors of simplices are contractible). The acyclic-models theorem then guarantees:

natural chain maps $F \to G$ and $G \to F$ extending the identity in degree $0$ exist and are unique up to natural chain homotopy;
any two such composites are naturally chain-homotopic to the identity.

So $A W$ and $\nabla$ , being two such natural maps that are the identity in degree $0$ , are forced to be chain-homotopy inverse. This is the prototypical acyclic-models argument and the technical foundation of the Künneth theorem and the cup product. May §29; the acyclic-models machinery of unit 03.12.34.

Exercise 8 (hard, proof). Dold-Kan is an equivalence: the unit and counit.

Sketch the proof that $N : sAb \to Ch_{\geq 0}$ and $Γ$ are inverse equivalences, by describing the natural isomorphisms $N Γ ≅ id$ and $Γ N ≅ id$ .

Hint

$Γ C$ in degree $n$ is $⨁_{[n] ↠ [k]} C_{k}$ . Show $N Γ C ≅ C$ by isolating the non-degenerate summand, and $Γ N A ≅ A$ via the normalization splitting.

Answer

The functor $Γ$ . For a chain complex $C$ , define $(Γ C)_{n} = ⨁_{[n] ↠ [k]} C_{k}$ , the sum over surjections in $Δ$ (equivalently, over iterated degeneracies). Face and degeneracy maps are defined by the simplicial-identity bookkeeping: degeneracies act by reindexing the surjections, and faces use the differential of $C$ on the "non-degenerate" summand.

$N Γ C ≅ C$ . The normalized complex picks out exactly the summand indexed by the identity surjection $[n] = [n]$ , namely $C_{n}$ , killing all properly-degenerate summands. The induced differential on this summand is the differential of $C$ (the $d_{0}$ on the normalized part reproduces $\partial^{C}$ ). So $N Γ C = C$ naturally.

$Γ N A ≅ A$ . By the normalization theorem (lead exercise), $A_{n} = ⨁_{[n] ↠ [k]} N_{k} A$ : every element of $A_{n}$ decomposes uniquely as a sum over its degeneracies of normalized elements (the splitting $C_{*} A = N_{*} A \oplus D_{*} A$ iterated). This decomposition is exactly the formula defining $(Γ N A)_{n}$ , and it is compatible with faces and degeneracies. So $Γ N A ≅ A$ naturally.

Both composites are the identity up to natural isomorphism, so $N$ and $Γ$ are inverse equivalences $sAb ≃ Ch_{\geq 0} (Ab)$ . The same argument works with $Ab$ replaced by any abelian category. May §22; Goerss-Jardine III.2.

Exercise 9 (hard, proof). Künneth via Eilenberg-Zilber.

Use the Eilenberg-Zilber theorem to derive the Künneth short exact sequence for $H_{*} (X \times Y)$ in terms of $H_{*} (X)$ and $H_{*} (Y)$ over a PID.

Hint

Eilenberg-Zilber gives $C_{*} (X \times Y) ≃ C_{*} (X) \otimes C_{*} (Y)$ ; apply the algebraic Künneth theorem for the tensor product of chain complexes.

Answer

By the Eilenberg-Zilber theorem (Exercise 7), there is a natural chain-homotopy equivalence $C_{*} (X \times Y) ≃ C_{*} (X) \otimes C_{*} (Y)$ , hence an isomorphism on homology:

H_{n} (X \times Y) ≅ H_{n} (C_{*} (X) \otimes C_{*} (Y)) .

Now apply the algebraic Künneth theorem for a tensor product of chain complexes of flat (e.g. free) modules over a PID $R$ : there is a natural short exact sequence

0 \to i + j = n ⨁ H_{i} (X) \otimes_{R} H_{j} (Y) \to H_{n} (X \times Y) \to i + j = n - 1 ⨁ Tor_{1}^{R} (H_{i} (X), H_{j} (Y)) \to 0,

and the sequence splits (non-naturally). The first term is the "naive" tensor product of homologies; the Tor term measures the failure of the tensor product of homology to compute the homology of the tensor product, exactly the torsion-interaction correction.

So the homological Künneth formula is the composite of two facts: Eilenberg-Zilber (a geometric/simplicial statement, proved by acyclic models) and the algebraic Künneth theorem (pure homological algebra over a PID). Over a field $R = k$ , Tor vanishes and one gets the clean $H_{n} (X \times Y) = ⨁_{i + j = n} H_{i} (X) \otimes H_{j} (Y)$ . May §29; Weibel §3.6.

Exercise pack EP5. May, Simplicial Objects in Algebraic Topology supplement: the Dold-Kan correspondence, simplicial abelian groups, the bar construction, and the Eilenberg-Zilber theorem across §17-§31.

Prerequisites

03.12.34 pending
03.12.25 pending
03.12.22 pending
03.12.41 pending
03.12.35 pending

Tier anchors

beginner
intermediate: May, Simplicial Objects in Algebraic Topology, §17-§23: the normalization theorem, the Dold-Kan correspondence, simplicial abelian groups, the bar construction, the Eilenberg-Zilber theorem
master

References

May — Simplicial Objects in Algebraic Topology · §17 (chain complexes from simplicial abelian groups, the normalized and Moore complexes), §22 (the Dold-Kan equivalence $\mathbf{sAb} \simeq \mathrm{Ch}_{\ge 0}$), §29-§31 (the bar construction and classifying spaces), §29 (the Eilenberg-Zilber theorem and the shuffle/Alexander-Whitney maps)
Goerss-Jardine — Simplicial Homotopy Theory · Ch. III §2 — the Dold-Kan correspondence and normalization
Weibel — An Introduction to Homological Algebra · §8.4 — the Dold-Kan theorem and simplicial methods in homological algebra

Estimated time

intermediate: 6h