By Botvinnik B.

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Then there exists a cell map g : X −→ Y such that g|A = f |A and, moreover, f ∼ g rel A. First of all, we should explain the notation f ∼ g rel A which we are using. Assume that we are given two maps f, g : X −→ Y such that f |A = g|A . A notation f ∼ g rel A means that there exists a homotopy ht : X −→ Y such that ht (a) does not depend on t for any a ∈ A. Certainly f ∼ g rel A implies f ∼ g , but f ∼ g does not imply f ∼ g rel A. 3. Give an example of a map f : [0, 1] −→ S 1 which is homotopic to a constant map, and, at the same time f is not homotopic to a constant map relatively to A = {0}∪{1} ⊂ I.

We use the map q : e(σ) −→ E(σ) to see that q(e(σ)) = E(σ). Thus we can describe π ∈ e(σ) \ e(σ) as a k -plane v1 , . . , vk , where σj vj ∈ H . Clearly vj ∈ Rσj , thus dim(Rσj ∩ π) ≥ j for each j = 1, . . , k . Hence τ1 ≤ σ1 , σj . , τk ≤ σk . However, at least one vector vj belongs to the subspace Rσj −1 = ∂ H , and k by corresponding τj < σj . Thus d(τ ) < d(σ). The number of all cells is equal to n counting. Now we count a number of cells of dimension r in the cell decomposition of G(n, k).

1, 2o . Indeed, let β = ηαǫ11 · · · ηαǫss , ǫj = ±1, where all elements ηα ηα−1 , ηα−1 ηα are canceled It is enough to show that β = e, where e is the identity elelement. Recall that ηα is given by the inclusion Sα1 −→ 1 α∈A Sα = XA . It is enough to construct a covering space p : T −→ XA so that the loop β is covered by a loop β with the property that β(0) = β(1). Consider s + 1 copies of the wedge XA placed over XA , see Fig. 5. We assume that these copies of XA project vertically on XA . Consider the word β = ηαǫ11 · · · ηαǫss .

### Algebraic topology notes by Botvinnik B.

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