M8130 Algebraic topology, tutorial 09, 2017
20. I 2017
Define the n-th homotopy group of the space X with the base point x0 as the group of homotopy classes of the cts1 maps (In,dln) —> (X,x0) with the operation given by prescription:
U     ' '    \g(2t1-l,t2,...,tn)   i <tt < 1.
Denote it nn(X,x0).
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 G I.
h(s, t\
Mi) Gl      (ti G [0,i + Si],sG [0,1])
<?M -
M^i-i^f)   Mi) e in (tiG[| + s|,i],se[o,i])
(?(4tx - (1 + s))    (s,t!)G ii    (*i G    + , „4
s±],sG [0,1])
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Exercise 2. 57jou;       the element given by prescription
(-/)(tl,...,t„) = /(l-tl,t2,...,tn)
is rea% £/«e 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).
h(s, t\
'l-s
Mi) e i
f(^) = (-f)m mi)gii
(-/)(2t1-l) = /(2t1)   Mi) Gill,
			
\	I	/	
j	V	n	
where ii = {(Mi) I s G [0,1],^ G [^, ±f*]).
Remark. One can see, that proving by pictures is much more pleasant.
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continuous
2 and some other following notations
M8130 Algebraic topology, tutorial 09, 2017
20. I 2017
There is a long exact sequence:
•••->• 7ln+1(X,A,x0) -> 7ln(A,x0) —7ln(X,x0) —7Vn(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, dln, Jn_1) -> (X,A,x0), where jn-i _ jn _        anc[ <9([/]) = = const, since / is constant on whole dln D /n_1.
"imj* D ker<9": Take an arbitrary g E kerd C 7rn(X, A, rr0), thus g: (In,dln, Jn_1) —)• (X, A, rr0). Since g E ker<9, there is the homotopy h: (In~1,dln~1) x J —>• (A,rr0) sucn that /i(rr, 0) = and h(x, 1) = const. Because h(x,t) G A and h(x',t) = xq for all
x G 7n_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) forte[0,±] h(x,2t-l) forte[±,l].
It is not hard to prove that j*(f) is homotopic to g, see picture below.
homotopic
Now, let us show the exactness in nn(A,x0).
"im<9 D keri*" Let / G keri* C 7rn(y4,x0) be an arbitrary. Because i*f ~ const, we have homotopy h: (In,dln) x/-> (X,x0) such that h(x,0) = f(x) and h(x, 1) = x0. It holds h G nn+1(X, A, x0), since /i(rr, 0) G A, h(x, 1) = rc0 and h(x',t) = x0 for all x E In and rr' G <9/n.
"im<9 C keri*" Let h E irn+i(X, A, xq) be an arbitrary. Denote h\in = f. Then h gives the homotopy i*f ~ const in (X, rro), since /i(rr, 0) = /(a?), h(x, 1) = rco and h(x',t) = xq 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 J
B
M8130 Algebraic topology, tutorial 09, 2017
20. I 2017
If p is a fibration then it has the homotopy lifting property also for all pairs (X, A) of CW- complexes:
X x {0} U A x I-s~ E
X x I-
Recall that p: E —>• B is a fiber bundle with fibre F if there are open subsets Ua such that B = [Ja Ua and the following diagram commutes for all Ua:
Ua x F ■
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}^-B x F
pr
Dn x I
B
Then h(—,0) = f and we can define H: Dn x I —> B x F by H(x,t) = (h(x, t), g(x)). One can see that the diagram commutes with H too.
Now, let p: E —> B be an arbitrary fibre bundle with fiber F and B = \Ja Ua. We can take In instead of Dn and consider a diagram:
In x {0}-
In x I
■E
B
Because In is compact, we can divide In x I to finitely many subcubes Ci x Ik where h = [jk,jk+i] such that h(Ci x Ik) C Ua for some a. Since Ua x F —^ B makes a trivial bundle, we can use the same approach as above for each subcube. Since we know -^|cjx{o} = /|cjx{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:
Ci x {0} UaxF
CiXl{
h
pr
-ua
M8130 Algebraic topology, tutorial 09, 2017
20. I 2017
Since we know H\
CiX{ji}
now, we can continue with the second "column":
Q x h
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 situation3:
In4
Co
Jo-0 TQ Ji     h h
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Exercise 5. Show the structure of the fibre bundle Sr'
lPn.
->• [x]. Now, we want to find a neigh-
Solution. The fibre is S° = { — 1,1}, since x, —x \-bourhood U of [x] such that p~l{U) = U x S°. Set U = {[x + v] \ v e [x]^} then we have homeomorphism <p: U x S° —> C Sn given by <p([x + v],l) = ^Xv\\ an<^
</?([rr + v], — 1) = jj^jji■ We can cover the whole IRPn by the open subsets U = {(x0 : xx :
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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, realise that we can consider S2n+1 C Cn+1. Take U0 = {[z0 : zx : ■ ■ ■ : zn] I 20 7^ 0} C CPn. We can consider U0 = {[1 : z\ : • • • : zn]}. Then the map f/oxS1^ S2n+1 is given by
[(1 : zx : ••• : ^e**] 1—>
{el\eltzx,.. .,eltzn) \(eit,eitz1,.. .,euzn)\
We can do the same for other U from the covering of IRPn.
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3it is drawn as planar, but it should be n-dimensional