ScalingStacks

Observe that, by excision, H3​(W′,W∪(Z∪L),ℤ)≅H3​(C,C′,ℤ)=0H_{3}(W^{\prime},W\cup(Z\cup L);{\mathbb{Z}})\cong H_{3}(C,C^{\prime};{\mathbb{Z}})=0. From here we obtain an exact sequence

(8) 0→H2​(W′,L,ℤ)→H2​(Z,∂−Z,ℤ)⊕H2​(W,ν⁡(K)∪L,ℤ)→H2​(C,C′,ℤ).0\to H_{2}(W^{\prime},L;{\mathbb{Z}})\to H_{2}(Z,\partial_{-}Z;{\mathbb{Z}})\oplus H_{2}(W,\nu(K)\cup L;{\mathbb{Z}})\to H_{2}(C,C^{\prime};{\mathbb{Z}}).

Thus, an element in H2L​(W′,ℤ)⊆H2​(W′,L,ℤ)H_{2}^{L}(W^{\prime};{\mathbb{Z}})\subseteq H_{2}(W^{\prime},L;{\mathbb{Z}}) can be identified with its image in H2​(Z,∂−Z,ℤ)⊕H2​(W,ν⁡(K)∪L,ℤ)H_{2}(Z,\partial_{-}Z;{\mathbb{Z}})\oplus H_{2}(W,\nu(K)\cup L;{\mathbb{Z}}), which we write as a pair (α,η)(\alpha,\eta).

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Ciprian Manolescu, Kevin Walker, Paul Wedrich

Original source: arXiv:2206.04616v2