M8130 Algebraic topology, tutorial 09, 2019
15. I 2019
Note: This tutorial originates in 20171.
Define the n-th homotopy group of the space X with the base point x0 as the group of homotopy classes of the continuous maps (In,dln) —> (X,x0) with the operation given by the prescription:
(f+ g)(t1,...,tn)
/(2ti,t2,... ,tn) g(2t1 - l,t2, ■ ■ ■ ,tr,
0 < ti < i \<h<\.
Denote it nn(X,xo). Note that a monoid structure induced by concatenation of maps with respect to other components as the first one coincides with above prescription by the Eckmann-Hilton argument. To see that / + g is a continuous map again we refer to the well-known pasting lemma.
Exercise 1. Show the operation on nn(X,x0) is associative.
Solution. We want to show (f + g) + k ~ f + (g + k). We will find prescription for the homotopy by the following diagram:
In the following notation2, understand        as f(t1,t2,..., tn) for all t2, ■ ■ ■ ,tn E I.
h(s, t\
(s,t1)El      (t! G [0,i + Si],sG [0,1])
H + s\],se [0,1])
0(4ti-(l + s))   (s,t1)Ell    (tiG[| + s
(2=1*1 - i^f)    MO e in (t1e[l + s4
□
Exercise 2. Show that the element given by prescription is really the inverse element of f.
Solution. We want to show / + (—/) ~ const. The constant will be (function given by) the point f(2t1) = /(0). Again let us draw a diagram (the square (its boundary) in the middle sign the same value; the wavy line sign one value too).
^ee https://is.muni.cz/el/sci/jaro2019/M8130/um/68045051/
2 and some other following notations
M8130 Algebraic topology, tutorial 09, 2019
15. I 2019
where II = {(s,£ij
/(2*i)
(-/)(l-2t1) = /(2t1)
Mi) G I Mi) G II (Mi) G III,
			
	I	7	
I	V	/ n	
□
Remark. One can see, that proving by pictures is much more pleasant. There is a long exact sequence:
•••->• 7ln+1(X,A,x0) 7ln(A,x0) 7ln(X,x0) 7ln(X,A,x0) 7Tn_i(A,X0) ->-...
Exercise 3. Show the exactness of this sequence in nn(X, A,x0) and nn(A,x0).
Solution. At first we will show the exactness in nn(X, A, x0).
Let us show the inclusion "imj* C ker<9". Take an arbitrary / G nn(X,x0), thus /: (In,dln) -> (X,x0). From definition = jo/: (Jn, <9/n, Jn_1) -> (X,A,x0), where jn-i _ jn _        anc[ <9([/]) = = const, since / is constant on whole dln D /n_1.
"imj* 3 ker<9": Take an arbitrary g G ker<9 C 7rn(X, A, x0), thus g: (In,dln, J'
n—1^
(X, A, rr0)- Since g G ker<9, there is the homotopy h: (Jn_1, <9Jn_1) x I —y (A,x0) such that h(x,0) = g\jn-i(x) and h(x, 1) = const. Because h(x,t) G A and h(x',t) = x0 for all x G Jn_1, x G <9/n_1 and £ G [0,1], we can take / G nn(X,x0) defined by
f(x,t)
g(x,2t) fortG[0,|] h(x,2t-l) fortG[|,l].
It is not hard to prove that j*(f) is homotopic to g, see picture below.
homotopic
Another approach to showing this inclusion is to view homotopy groups nn(X, A, x0) as the homotopy classes [(Dn, S'^1, s0), (X, A, x0)] and the connecting homomorphism d as the restriction Is^-i. Then we have the following commutative diagram
Dn x {0} U S"-1 x I
H ~
gUh
X
Dn x I
M8130 Algebraic topology, tutorial 09, 2019
15. I 2019
The homotopy H exists by HEP of the pair (Dn, S'"1-1). Then we can take [H(-, 1)] G TTn(X,x0) and j*([H(-, 1)]) = [g].
Now, let us show the exactness in nn(A,x0).
"im<9 D keri*" Let / G keri* C 7rn(y4,xo) be an arbitrary. Because i*f ~ const, we have homotopy /i: (In,dln) x/-> (X, rr0) such that /i(rr, 0) = /(rr) and h(x, 1) = rr0- It holds h E 7rn+i(X, A, xq), since /i(rr, 0) G A, h(x, 1) = rc0 and h(x',i) = x0 for all x E In and rr' G dln.
"im<9 C keri*" Let h E irn+i(X, A,x0) be an arbitrary. Denote h\jn = f. Then h gives the homotopy i*f ~ const in (X, rr0), since /i(rr, 0) = /(a?), h(x, 1) = rc0 and h(x',t) = x0 for all x' E dln and x E In. □
^4 map p: E ^ B is called a fibration if it has the homotopy lifting property for all (Dn,®):
Dn x {0}-E
Dn x I
B
If p is a fibration then it has the homotopy lifting property also for all pairs (X, A) of CW- complexes:
X x {0} Li Ax I-E
X x I-^B
Recall that p: E —>• B is a fibre bundle with fibre F if there are open subsets Ua such that B = [Ja Ua and the following diagram commutes for all Ua:
Un x F ■
homeo
Typical examples of fibre bundles with fibre S1 over I are a trivial bundle or Mobius band?. Exercise 4. Show that every fibre bundle is a fibration.
Solution. At first consider a trivial fibre bundle E = BxF. Take an arbitrary commutative diagram of the form:
Dn x {0}
Dn x I
BxF ^B
3http://rin.io/intro-to-bundles/
M8130 Algebraic topology, tutorial 09, 2019
15. 4. 2019
Then h(-,0) = f and we can define H: Dn x I -> B x F by F(M) = £), prF(/(x))). One can see that the diagram commutes with H too.
Now, let p: E —> B be an arbitrary fibre bundle with fibre F and B = \Ja Ua. We can take In instead of Dn and consider a diagram:
r x {o}^~£
In X I
■B
Because In is compact, we can divide In x I to finitely many subcubes Cj x Ik where h = [jk,jk+i] such that h(Ci x Jfc) C £/"a for some a. This is possible due to Lebesgue's number lemma. Since Ua x F ^ Ua makes a trivial bundle, we can use the same approach as above for each subcube. Since we know H\CiX{o}, we can find the lift H for all cubes in the first "column" (see the picture below) in the same way as for the trivial case (as we know that each Serre fibration has HLP w.r.t. a pair of CW-complexes)
d x {0} U A® x J0 Q x I0-
H ^ -
h
Un x F
pr
where A\ = U^ncOx/^ciw^^ n Since we know H\c^{h] now, we can continue with the second "column":
C,x{3l}UA\xh
/UHI
ft
rja x f
pr
Thus, we can proceed through all columns in this way until we will get H on the whole In x I. The illustration of this situation4:
in4
Jo-0 JQ Ji     h h
4it is drawn as planar, but it should be n-dimensional
M8130 Algebraic topology, tutorial 09, 2019
15. I 2019
□
Exercise 5. Show the structure of the fibre bundle Sn IRPn.
Solution. The fibre is S° = { — 1,1}, since x, —x i—> [x]. Now, we want to find a neighbourhood U of [x] such that p~l{U) = U x S°. Assume that
[x] = [1 : xx : • • • : xn] E U0 = {[y0 ■ Vi ■ ■ ■ ■ ■ yn] \ Vo 0}
then we have the homeomorphism <p: U0 x S° —> P_1(^o) ^ Sn given by </?([rr],l) = £'Xl'"''Xn\H and u?(\x],-l) = i~H),~Xl,-~^)•  We can cover the whole Rpn by the °Pen
\\(l,Xi,...,Xn)\\ -1' I J r
subsets Ui = {[y0 : y± : ■ ■ ■ : yn] \ Vi ^ 0}. □ Exercise 6. Show the structure of the fibre bundle S2n+1       CPn with the fibre S1.
Solution. Let us look on the special case S1 ^ S3 —> CP1 = S2 called "Hopf fibration". Realise that we can consider S3 C C2, so we can (locally) define the projection S3 —> CP1
by (21,2:2) ^ %■
In the general case, realize that S2n+1 C Cn+1. Take U0 = {[z0 : zx : • • • : zn] \ z0 ^ 0} C CPn. Then the map U0 x S1 -> p-l{UQ) C S2n+1 is given by
( /O^t pit   y pit   y \
(eit,eitz1,... ,euzn)
We can do the same for other U from the covering of CPn. □