By Iain T. Adamson

This paintings goals to provide basic topology in an unconventional manner. It supplies a evaluate of the elemental definitions including workouts with out recommendations or proofs of the theorems partially 1, after which offers the ideas partially 2, permitting the coed to check solutions with their very own.

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**Example text**

The map d is a boundary operator: the composition d2 is δ◦ ρ ◦ δ ◦ ◦ δ, but δ : Hn−1 (X n−1 ) → Hn−1 (X n−1 , X n−2 ) → Hn−2 (X n−2 ) is a composition in a long exact sequence and thus vanishes. 5 Let X be a CW complex. The cellular chain complex C∗ (X) consists of the free abelian groups Cn (X) := Hn (X n , X n−1 ) with boundary operator d : Hn (X n , X n−1 ) where δ GH n−1 (X n−1 ) GH n−1 (X n−1 , X n−2 ) is the map induced by the projection map Sn−1 (X n−1 ) → Sn−1 (X n−1 , X n−2 ). 6 (Comparison of cellular and singular homology).

Thus, ∂B k (e) − ∂e = ∂ ψ˜n−1 (c) c = ∂e = ∂(B k (e) − ψ˜n−1 (e)) is a boundary in SnU (X) as well. We remark that this isomorphism actually comes from a homotopy equivalence of chain complexes. 14 (Excision). ˉ ⊂ A. ˚ Then the inclusion i : (X \ W, A \ W ) → (X, A) induces Let W ⊂ A ⊂ X such that W an isomorphism Hn (i) : Hn (X \ W, A \ W ) ∼ = Hn (X, A) for all n 0. Proof. • We first prove that Hn (i) is surjective. e. ∂c ∈ Sn−1 (A). Consider the open covering ˚X\W ˉ } =: {U, V } of X. Now subdivide and find k such that c := B k c is a U = {A, U chain in Sn (X).

Note that for any topological space X, the cone CX is contractible to its apex. Thus ˜ n (CX, CA) = 0 ˜ n (CX) = 0 for all n 0. Similarly, for A ⊂ X, we have CA ⊂ CX and H H for all n 0. 3. The suspension of Sn is ΣSn ∼ = Sn+1 . 4. We have natural embeddings X → CX and CX → ΣX. We can see the suspension as two cones, glued together at their bases. 11 (Suspension isomorphism). Let A ⊂ X be a closed subspace and assume that A is a deformation retract of an open neighbourhood A ⊂ U . Then ˜ n−1 (X, A), Hn (ΣX, ΣA) ∼ =H for all n > 0.