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If there are neighbourhoods Ui of xi ∈ Xi together with a deformation of Ui to {xi }, then we have for any finite E ⊂ I Hn (Xi ). Hn ( Xi ) ∼ = i∈E i∈E In the situation above, we say that the space Xi is well-pointed with respect to the point xi ∈ Xi . Proof. First we consider the case of two bouquet summands. We have X1 ∨ U2 ∪ U1 ∨ X2 as an open covering of X1 ∨ X2 . Since (X1 ∨ U2 ) ∩ (U1 ∨ X2 ) = U1 ∩ U2 is contractible, the Mayer-Vietoris sequence then gives that Hn (X) ∼ = Hn (X1 ∨ U2 ) ⊕ Hn (U1 ∨ X2 ) for n > 0.

Indeed, if n > 0 would be the lowest dimension of a cell, Sn−1 could not be taken into cells of dimension at most n − 1. 6. It follows from axiom (a) that for every n-cell σ, we have σ = Φσ (Dn ). From the general inclusion f (B) ⊂ f (B) for continuous maps, we conclude Dn ) ⊃ Φσ (Dn ) ⊃ σ . σ = Φσ (˚ As a compact subspace of a Hausdorff space, Φ σ (Dn ) is closed; since it lies between σ and σ, we conclude Φσ (Dn ) = σ. In particular the closure σ is compact in X as the continuous image of the compact set Dn .

11. 1. Arbitrary intersections and arbitrary unions of subcomplexes are again subcomplexes. 2. The skeleton X n is a subcomplex. 3. Every union of n-cells in X with X n−1 forms a subcomplex. 4. Every cell lies in a finite subcomplex. Proof. 1. 10 a subcomplex. The statement about the union follows from the definition of a subcomplex. 2. and 3. follow from the observation that for an n-cell σ we have that σ = (σ \ σ) ∪ σ is contained in X n−1 ∪ σ. 4. Induction on the dimension of the cell; then use closure finiteness and σ = Φσ (Dn ).

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