Solutions to Problem Set 1

Home , Cone (topology), Mapping cone (topology)

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 ∈ Y and (x , t) 7→ (φ(x ), t) for (x , t) ∈ 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 ψ : 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→ (ψ(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) ∈ Y t ≤ 1 (x, t) 7→ 2 1 (x, 2t − 1) ∈ 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) ∈ Y × I and  s fg(x, 2t) ∈ Y t ≤ 2 (x, t, s) 7→ 2 s s  g(x, s), 2−s t − 2−s ∈ X × I 2 ≤ t

for (x, t, s) ∈ (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 {x0} ∈ 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 ∈ X. Consider now the map h = r ◦ h : A × I → A. For a ∈ 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 {r(x0)} ⊂ A, and hence A is contractible (again by [Bredon, Prop. I.14.5]).

3. Recall that Sn = {x ∈ Rn+1 kxk = 1}. 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 ∈ 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) ∈ 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 × {−1} with each other, and all points of S1 × {+1} 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 ∈ 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 ∈ [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] ∈ 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 × {1} to a single point, namely [0] ∈ RP , it descends further to a map F : Cf → RP . 00 0 1 Note that F is bijective, since F is injective on Mf \ (S × {1}). 2 Since Cf is compact and RP is Hausdorff, we conclude using [Bredon, Theorem I.7.8] that φ is a homeomorphism.