FIRST ASSIGNMENT. (1) Let E X X be an equivalence relation on a set X. Construct the set of equivalence classes as colimit in the category Sets.

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1 FIRST SSIGNMENT DUE MOND, SEPTEMER 19 (1) Let E be an equivalence relation on a set. onstruct the set of equivalence classes as colimit in the category Sets. Solution. Let = {[x] x } be the set of equivalence classes and q : be the quotient. Let p 1, p 2 : E be the two projections. e show that E p 1 p 2 q is a coequilizer diagram. Then is isomorphic to the set of equivalence classes. map satisfies f(p 1 (x, y)) = f(p 2 (x, y)) if and only if f(x) = f(y) if (x, y) E. Define f : by f ([x]) = f(y) for any choice of y [x]. This is well defined since y [x] implies that (x, y) E so f(x) = f(y). Further, f (q(x)) = f ([x]) = f(x) so f q = f. Let f : be any other map such that f q = f. Then, for [x], f ([x]) = f(q(x)) = f ([x]) so f = f and the hence f is unique defined. Therefore, (2) Let be a commutative diagram. (a) Prove that if the two inner squares are pushouts, then so is the outer rectangle. That is, suppose that both = and =. Prove that =. Solution. Given maps and making 1

2 2 DUE MOND, SEPTEMER 19 commute, since the left hand square is a pull back, we get a unique map making the diagram commute. Now, since the right hand square is a pull-back, we obtain a unique map making the diagram commute. Therefore, has the required universal property. (b) hat about if = and =, then is =? nd what if = and =, is =? Solution. If = and =, then =. Indeed, suppose that we have maps making the right hand diagram commute. Then the composites and agree, and since the outer rectangle is a pushout, we obtain a

3 FIRST SSIGNMENT 3 unique map f such that is equal to and is equal to. e need to prove that is equal to. However, both these maps fulfill the universal property for the diagram for the maps and f =. Since is the pushout, the two maps must be equal. On the other hand, if = and =, then it is not necessarily the case that =. n counter-example in sets is given by {1} {1, 2} {1} {1} {1} {1} where the maps are the inclusions when there is a choice. (3) In the following problem, let S n 1 D n be the inclusion of the boundary, be the map (id ) ( id) and be the fold map id id. For a based topological space, let J 2 () = ( )/((x, ) (, x)). That is, J 2 () is the push-out J 2 ().

4 4 DUE MOND, SEPTEMER 19 Further, you may use the fact that D 2n = I 2n = D n D n and other such standard homeomorphisms without proof. (a) Describe S n S n as a -complex obtained from S n S n by attaching a single 2n cell. Exhibit this as a pushout. Solution. e first look at the more general situation where we have and with quotient maps both denoted by q : / and q : /. e prove that (q ) ( q) / / q q (1 ) ( 1) / / F G is a pushout. Suppose we are given maps F : and G : / / making the diagram commute. Define H : / / by H(x, y) = G(x, y). Then, for any a, H(a, y) = G(a, y) = F (y) and for any b, H(x, b) = G(x, b) = F (x) so this is well-defined. It is also continuous since for any open U, H 1 (U) = (q q)(g 1 (U)) Further, this map is uniquely defined since q q is surjective.

5 FIRST SSIGNMENT 5 Now, consider (, ) = (, ) = (D n, D n ). pushout The previous construction gives a D n D n D n D n D n / D n D n / D n D n D n D n / D n D n / D n. Identifying (D n D n ) = D n D n D n D n, D n / D n = S n and D n D n = D 2n proves the claim. (b) Use your construction in (a) to give J 2 (S n ) the structure of a -complex with one n cell and one 2n cell. Solution. onsider the following commutative diagram, where the right hand square is the pushout defining J 2 (S n ) and the left hand square is the pushout of part (a). D 2n S n S n S n D 2n S n S n J 2 (S n ). y 2(a), the outer square is a pushout diagram, which proves the claim. (c) Show that S n is an H-space if and only if the attaching map of the 2n-cell of J 2 (S n ) is null-homotopic. Solution. First, note that if S n S n µ S n gives S n the structure of an H space, then there are homotopies H : S n I S n from µ (1 ) to the identity and H : S n I S n from µ ( 1) to the identity. Letting H : (S n S n ) I S n be given by H (x, t) if x is in the left factor and H (x, t) if it is in the right factor, we obtain a homotopy from µ ((1 ) ( 1)) to the fold map : S n S n S n. Therefore, if S n is an H space, then S n is homotopic to (1) (1 ) ( 1) S n S n S n.

6 6 DUE MOND, SEPTEMER 19 However, since (1 ) ( 1) S n S n is equal to D 2n D 2n S n S n, (1) extends to D 2n, and therefore is null-homotopic. onversely, suppose that S n is null-homotopic. Then it extends to a map D 2n S n making the following diagram commute D 2n D 2n S n S n S n S n Since the square is a pushout, we get a lift S n S n S n. Further, by the commutativity of the right triangle, this gives S n the structure of an H space. S n (4) Let be 1 connected (i.e., π 0 = π 1 = 0). Recall that for a covering map p : E of based spaces, given a base point preserving map f :, there exists a unique base point preserving lift f : E such that p f = f. (a) Prove that π n p : π n E π n is an isomorphism for n 2. Solution. Note that for n 2, S n and S n I are both 1 connected. e need to prove that π n p is injective and surjective. Since any map f : S n lifts uniquely to a map f : S n E such that p f = f, π n p is surjective. Let f, g : S n E be such that [p f] = [p g]. Let h : S n I be a homotopy between p f and p g. Then there exists a unique lift H : S n I E such that p H = h. In particular, p H 0 = p f and p H 1 = p g. Therefore, since H 0 lifts p f but so does f, we have H 0 = f and similarly, H 1 = g. Therefore, f g and [f] = [g] so that π n p is injective. (b) ompute π k RP n in terms of π k S n. hat about π k RP? Solution. If n = 1, RP 1 = S 1 so π k S 1 = π k RP n. Suppose that n 2. Then let p : S n RP n be the antipodal map which is the quotient of S n by the equivalence relation x x. This is a covering space. In fact, since S n is simply connected, it is

7 FIRST SSIGNMENT 7 the universal cover of S n. Since p 1 ( ) = 2, π 1 RP n is /2. For n 2, the previous problem implies that π k RP n = π k S n. Recall that RP = n RP n. Define S = n=1 Sn with the union topology. Let p : S RP defined by the union of the map p : S n RP n. Then p : S RP is a covering map. s before, π 1 RP = /2. Note that S is contractible, so π k RP = 0 for n 2.