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Solutions to Problem Set 1 Prof. Paul Biran Algebraic Topology I HS 2013 ETH Z¨urich Solutions to problem set 1 Notation. I := [0; 1]; we omit ◦ in compositions: fg := f ◦ g. 1. It follows straight from the definitions of \mapping cone" and \mapping cylinder" that the 0 0 0 0 map F~ : Mfφ ! Mf given by y 7! y for y 2 Y and (x ; t) 7! (φ(x ); t) for (x ; t) 2 X × I is well-defined and descends to a map F : Cfφ ! Cf . We will show that F~ is a homotopy equivalence, which implies that the same is true for F . Let : X ! X0 be a homotopy inverse of φ, and let g : X × I ! X be a homotopy connecting φψ to idX . Consider the map G~ : Mfφψ ! Mfφ defined analogously to F~; that is, it maps y 7! y and (x; t) 7! ( (x0); t). The composition F~G~ : Mfφψ ! Mf is then given by y 7! y and (x; t) 7! (φψ(x); t). Note that we have another map h : Mfφψ ! Mf induced by the homotopy fg : X × I ! Y connecting fφψ to f: h is given by y 7! y and ( fg(x; 2t) 2 Y t ≤ 1 (x; t) 7! 2 1 (x; 2t − 1) 2 X × I 2 ≤ t and [Bredon, Theorem I.14.18] tells us that h is a homotopy equivalence. We will now show that F~G~ ' h. Consider the homotopy Σ : Mfφψ × I ! Mf given by (y; s) 7! y for (y; s) 2 Y × I and 8 s <fg(x; 2t) 2 Y t ≤ 2 (x; t; s) 7! 2 s s : g(x; s); 2−s t − 2−s 2 X × I 2 ≤ t for (x; t; s) 2 (X × I) × I. Note that Σ is well-defined and continuous and satisfies Σ(·; 0) = F~G~ and Σ(·; 1) = h. It follows that F~G~ is a homotopy equivalence, too. In an analogous manner, one can define a map H~ : Mfφψφ ! Mfφψ and then show that G~H~ : Mfφψφ ! Mfφψ is a homotopy equivalence. The rest of the proof is analogous to the end of the proof of [Bredon, Theorem I.14.19]. Namely, let k and k0 be homotopy inverses of F~G~ resp. G~H~ , so that in particular kF~G~ ' ~ ~ 0 ~ ~ 0 idMfφψ and GHk ' idMf . Then L := kF and R := Hk are left resp. right homotopy inverses of G~. By Problem 4 on this sheet, L and R are in fact homotopy inverses of G~, and hence G~ is a homotopy equivalence. Hence F~G~ and G~ are both homotopy equivalences, from which it follows that also F~ is a homotopy equivalence. 2. By [Bredon, Prop. I.14.5], contractibility of X means that idX is homotopic to a map whose image is a singleton fx0g 2 X; that is, there exists a map h : X × I ! X such that 0 h(x; 0) = x and h(x; 1) = x0 for all x 2 X. Consider now the map h = r ◦ h : A × I ! A. For a 2 A, it satisfies h0(a; 0) = r(h(a; 0)) = r(a) = a by the defining property of retraction, 0 and h (a; 1) = r(x0) ; in other words, idA is homotopic to a map with image fr(x0)g ⊂ A, and hence A is contractible (again by [Bredon, Prop. I.14.5]). 1 3. Recall that Sn = fx 2 Rn+1 kxk = 1g. We claim that the expression (1 − t)f(x) + tg(x) (x; t) 7! ; k(1 − t)f(x) + tg(x)k yields a well-defined map φ : X × I ! Sn. Pretending that this is true, note that φ(x; 0) = f(x) and φ(x; 1) = g(x) for all x 2 X, and hence f and g are homotopic. To prove well-definedness, we must show that the denominator never vanishes. To do so, we assume the contrary, i.e., that there exists some (x; t) 2 X×I such that (1−t)f(x)+tg(x) = 0. This is equivalent to (1 − t)f(x) = −tg(x), from which we obtain (1 − t) · kf(x)k = t · kg(x)k; 1 since kf(x)k = kg(x)k = 1, it follows that 1 − t = t, and hence t = 2 . Inserting this into the first equation, we obtain f(x) = −g(x), which contradicts our assumption. 4. Using the assumptions fg ' idY and hf ' idX , we obtain fh = fh idY ' fhfg ' f idX g = fg ' idY : Hence h is in fact a homotopy inverse of f, and thus f is a homotopy equivalence. (Similarly, one can show that g is a homotopy inverse of f.) 5. One could prove these statements by writing down explicit maps and homotopies. To facil- itate this, one could use, for example, the following \model" for S2 (and similar models for X; V; S2 _ S1 and S2 _ S2). The map 1 2 p 2 p 2 S × [−1; 1] ! S ; (x1; x2; h) 7! 1 − h x1; 1 − h x2; h ; induces a homeomorphism (S1 × [−1; 1])= ∼ ≈ S2; where ∼ is the equivalence relation that identifies all points of S1 × {−1g with each other, and all points of S1 × f+1g with each other. (Namely, the map is continuous and bijective, and since S1 × [−1; 1] is compact, it is a homeomorphism.) Another less direct way of proving the required statements is to use the following fact: If (Z; A) is a CW pair consisting of a CW complex Z and a contractible subcomplex A, then the quotient map Z ! X=A is a homotopy equivalence. (See [Hatcher p.11].) To use this, one must endow the spaces X and Y with appropriate structures of CW complexes. For X, consider an additional arc A on S2 connecting the north and the south pole; a possible CW complex structure has the two poles as 0-cells, the interiors of the \stick" and the arc A as 1-cells, and the rest of S2 as a 2-cell. The statement above implies that X ! X=A is a homotopy equivalence, but also that X=A is homotopy equivalent to S2 _ S1 (namely, denote by X~ the CW complex obtained by removing the stick, whose associated space is S2; then X=A~ is homotopy equivalent to S2 by the statement above; from this it follows easily that X=A is homotopy equivalent to S2 _ S1). For Y , consider the CW complex structure with one 0-cell on the equator, one 1-cell (the rest of the equator), and three 2-cells (northern and southern hemispheres and the disc in the equator plane). Denoting the last-mentioned 2-cell by A, the statement above tells us that Y ! Y=A is a homotopy equivalence; on the other hand, it is easy to see (using e.g. the model above), that Y=A is homeomorphic to S2 _ S2. 6. We view RP 2 as D2= ∼, where ∼ is the equivalence relation that identifies antipodal points on the boundary. That is, x; y 2 D2 satisfy x ∼ y if and only if x = y, or x and y both lie on S1 ⊂ D2 and satisfy x = −y. 2 Consider the map F : S1 t (S1 × I) ! RP 2 defined on S1 by e2πit 7! [eπit] for t 2 [0; 1], and on S1 × I by (e2πit; s) 7! [(1 − s)e2πit]. (Note that the first part is well-defined and continuous, because e2πi0 = −1 ∼ 1 = e2πi1 and hence [e2πi0] = [e2πi1] 2 RP 2.) 0 2 2πit Observe that F descends to a well-defined map F : Mf ! RP , because F (f(e )) = F (e4πit) = [e2πit] = F (e2πit; 0), and moreover F 0 is clearly surjective. Since F 0 maps all of 1 2 00 2 S × f1g to a single point, namely [0] 2 RP , it descends further to a map F : Cf ! RP . 00 0 1 Note that F is bijective, since F is injective on Mf n (S × f1g). 2 Since Cf is compact and RP is Hausdorff, we conclude using [Bredon, Theorem I.7.8] that φ is a homeomorphism. 3.
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