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Stable unoriented Thom homotopy is injectively detected

Statement

Assume AC. For each n≥0, the stable group π_n(MO)=colim_r π_{r+n}(T_r) maps injectively to V_n=F₂^{B_n} by the compatible detector coordinates D_{r,n}, using the cofinal tail r≥n+2.

Facts & Assumptions

Given: AC; a stable degree $n \geq 0$ ; the stable group $π_{n} (M O) = {colim}_{r} π_{r + n} (T_{r})$ over the cofinal tail $r \geq n + 2$ ; the compatible detector coordinates $D_{r, n}$ of the suspension-compatibility lemma; and the target $V_{n} = F_{2}^{B_{n}}$ .

[F1]

The finite-range theorem makes each $D_{r, n}$ an isomorphism for $r \geq n + 2$ , and the suspension-compatibility lemma gives $D_{r + 1, n} \circ (β_{r})_{*} = D_{r, n}$ with identity target bonding maps (The finite Thom detector is a homotopy isomorphism through 2r−2, Stable Thom detector coordinates commute with suspension).

[F2]

The stable homotopy colimit is computed over any cofinal tail (Stable homotopy groups of a sequential prespectrum, Stable homotopy colimits are independent of a cofinal tail); AC fixes the global basis defining the coordinates (The Axiom of Choice).

Proof

technique · direct

1.1givenF1

With the fixed target $V_{n}$ , the target bonding maps are identities, and the finite-range theorem makes each $D_{r, n}$ an isomorphism on every rank $r \geq n + 2$ . The cofinal-tail lemma then yields injectivity of the colimit map.

2.1step 1.1F1F2∎

Explicitly, represent a stable class at a rank r≥n+2. If its detector is zero, compatibility makes its detector zero at every later rank; injectivity of D_{s,n} then makes the advanced source representative zero, so the colimit class was zero. This proves the stable conclusion without assuming stabilization is an isomorphism in advance. In fact, because each D_{r,n} is an isomorphism and the square commutes, the bonding maps are isomorphisms on this tail, but this stronger consequence is not needed for injectivity.

Depends on

Used by

Thom's theorem: Stiefel-Whitney numbers detect unoriented bordism Theorem

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel S. Freed, Bordism: Old and New (standard reference, not scraped)