Topology Hmwk 3
All problems are from Allen Hatcher Algebraic Topology (online) ch 1
Andrew Ma
December 23, 2013
This assignment has been corrected post - grading.
1.1.16
(a)
Proof. Assume for a contradiction that there exists a retract. By prop 1.1.17 a retract
implies that the homomorphism induced by the inclusion of A ,→ X is injective. This
implies π1 ( A) ,→ π1 ( X ), or Z injects into the trivial group which is a contradiction.
(b)
Proof. Similarly apply prop 1.1.17 for a contradiction. If a retract existed then this would
imply an injection from π1 (S1 × D2 ) ∼
= π1 ( S1 ) × π1 ( D 2 ) ∼
= Z to π1 (S1 × S1 ) ∼
= Z × Z,
which is impossible.
(c)
Proof. Note that the space A is a deformation retract of a space, I’ll call T the twisted
torus (essentially a torus with a twist in it - I’ll try to remember to attach a picture
of this space). Then see that T also deformation retracts to something homeomorphic
to S1 ∨ S1 therefore A is homotopically equivalent to S1 ∨ S1 so that the fundamental
groups are isomorphic. That is π1 ( A) ∼
= π1 (S1 ∨ S1 ). Then see that π1 (S1 × D2 ) ∼
= Z.
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The as in previous reasoning pi1 (S ∨ S1 ) does not inject into Z.
(d)
Proof. X clearly deformation retracts to a point so that π1 ( D2 ∨ D2 ) ∼
= 0. It is also clear
1 ∨ S1 contains loops which aren’t homotopic to a point.
that π1 (S1 ∨ S1 ) 6∼
0
because
S
=
Thus by prop 1.1.17 there can’t exist a retract because π1 (S1 ∨ S1 ) 6,→ 0.
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(e)
Proof. X has a clear deformation retract to a circle therefore π1 ( X ) ∼
= π1 ( S1 ) ∼
= Z.
1
1
∼
However π1 (S ∨ S ) = Z ∗ Z (this is from chapter 1.2 with Van Kampen’s thm but
even if you didn’t have this you could probably show that the group π1 (S1 ∨ S1 ) isn’t
generated by just one loop - perhaps by showing it directly). At any rate there can’t be
a retract because π1 (S1 ∨ S1 ) does not inject into Z.
(f)
Proof. The Mobius band can deformation retract to a circle so π1 ( X ) ∼
= π1 ( S1 ) ∼
= Z.
1
1
The boundary circle of the Mobius band is homeomorphic to S ∨ S so that π1 ( A) ∼
=
π1 (S1 ∨ S1 ), by reasoning similar to that in problem (d) we see that pi1 ( A) 6,→ π1 ( X ) so
that there cannot be a retract from X to A.
1 1.1.17
Proof. Let the map rn : S1 ∨ S1 → S1 be a map which fixes one copy of S1 but maps
the other copy S1 to n loops in the first copy of S1 based at the wedge point. This map
is will be a retraction to the first copy of S1 since this is a cts map and it fixes the first
copy of S1 . Finally see that rn 6' rm if n 6= m. We can see this is true by restricting these
maps to the copy of S1 which is sent to n loops or m loops respectively. There can be no
homotopy between these maps and it is enough to show this since there is an obvious
homotopy between the maps rn and rm when restricted to the copy of S1 which is fixed,
then by cancellation (which we have shown for path homotopies, but I believe works
in this scenario too), this induces a homotopy of rn and rm when restricted to the copy
of S1 that gets mapped to loops. To see that no such induced homotopy can exists, note
that any cts map from S1 to S1 is a loop in the space S1 . Without loss of generality we
can assume that these loops are based at the wedge point (because S1 , which we are
mapping to, is path connected). Then a homotopy of the restricted maps would be a
path homotopy of an n loop to an m loop which can’t exist if n 6= m. Thus each rn is
nonhomotopic to another rm when n 6= m and therefore we get an infinite number of
retracts.
1.2.5
**Note to self : I believe that I could prove a recursive algorithm to show that given a
set of generating loops any loop could be created from the generators, however I did
not have time to complete this argument **
I got the solution to this problem from a post on math stack exchange. Here is a link
for reference
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http://math.stackexchange.com/questions/179064/good-exercises-to-do-examples-to-illustrat
because I am going to type up the general idea of this proof very briefly.
Proof. In the graph X we can take a maximum spanning tree, M. If M = X then we
are done because the fundamental group of a tree is a trivial group. If M 6= X then
label the finite number of edges in X \ M as e1 , . . . , en . Consider a cover of X by sets
M ∪ ei for i = 1 . . . n. (I have yet to see why these sets are open - but I’m running only a
few minutes before class). I believe that we can make the M ∪ ei open by taking slightly
larger open endpoints of this set. The triple intersection and pairwise intersections both
contain M so they are path connected. We may now apply Van Kampen’s thm. The
pairwise intersection of each set M ∪ ei ∩ M ∪ e j for i 6= j is simply M and therefore is
trivial. The normal group containing the inclusions of these trivial fundamental groups
is therefore trivial. Therefore, the fundamental group of X is simply ∗n π1 ( M ∪ ei ). All
that is left is to compute the π1 ( M ∪ ei ) for each ei . This fundamental group can be
generated by a loop starting from a fixed base point, x0 ∈ X, which travels once around
the bounded region that has edge ei . This shows the claim.
1.2.6
I don’t know how to show the last part of this problem for dimension equal to 3.
Claim: The inclusion X ,→ Y induces an isomorphism on π1
Proof. Without going into too much detail I believe that we can repeat the argument
given in the proof of Prop 1.26 using n-cells instead of 2-cells. The only difference
would be noticing that the π1 ( A ∩ B) = 0 because each open set Aα (used to apply Van
Kampen’s thm to A ∩ B) can deformation retract to a point due to the n-cell, being an
n dimensional closed ball, with a point removed is still contractible (because n ≥ 3).
Thus we get that the injection of fundamental groups is actually an isomorphism.
**Note to self : In showing that the space has trivial fundamental group I don’t
why we would use CW complexes when you can produce a wedge sum and solve the
problem in similar way to determining π1 of the wedge of circles**
Claim: The complement of a discrete subspace of Rn is simply connected for n ≥ 3
Proof. Let A be the discrete subspace in Rn (for n ≥ 3). First I will show that Rn \ A is
path connected.
Since A is discrete, for each pt a in A, there exists an open ball in Rn s.t. this ball
intersects with A contains only a, call this ball Bα . Then consider the collection of balls
in Rn . Note the fact that the boundaries of these n-dimensional balls are path connected
(this is shown in Munkres example 5 on page 156 - Using a cts map from Rn \ {0} →
Sn−1 ).
Now consider any two points of Rn \ A. If these two points are not contained in any
ball Bα , then draw a straight line between the two points and if this line intersects the
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boundary of any ball Bα then use the path connectedness of the boundary to connect one
point of the intersection to the opposite point of intersection instead of going straight
through the ball. Doing this for each ball the line meets will result in creating a new
path which still connects the two points in question.
If the two points are each contained in balls. For each point take a straight line path
from to the boundary of each respective ball and finally connect the boundaries of each
ball by a straight line in a similar manner to the above. In fact using this same idea we
can produce paths between pairs of points in any case i.e. one point is in a ball the other
is not in a ball, both points are in the same ball, etc.
Now I will argue that the space Rn \ { A} has trivial fundamental group. Given that
for each pt a ∈ A there is a nbhd of a which contains no other points of A, I’ll call this
nbhd Bα . We can deformation retract the space Rn \ A to a collection of boundaries of
balls (one for each Bα ) each of which is connected by a path as described in the previous
paragraph to the origin. That is each is connected by a straight line to the origin which
”paths” around any boundary of a sphere which it intersects. (Aside : The retract may
be done by considering a smaller closed ball inside each of the open balls Bα )
*At first I thought the ability to perform this deformation retract was clear by thinking of examples in R3 of balls, however I realize now it may not be so clear, but it is too
late in the night for me to turn back*.
Assuming that this deformation retract can be done we can then consider these boundaries of balls as copies of en−1 cells which have been attached to a collection of line segments (these line segments are the paths connecting the boundary of each ball to the
origin. The fundamental group of this collection paths is trivial. If n > 3 then by the
earlier part of 6 the attaching the en−1 cells do not change the fundamental group.
In the case of n = 3 we can say that . . . I guess I’m not sure. Hunh?
1.2.8
We can be more specific in this problem. There is a more detailed answer at www3.nd.edu/ Inicolae/ProblemsHatcher.pdf.
Proof. Consider the attachment of two torii seen as two labeled squares with the attached along a side. Say the edges are labeled a, b, c where the squares are attached
along edge a. Then by applying Van Kampen’s theorem for CW - complexes (this space
is a CW - Complex) the resulting fundamental group is < a, b, c| aba−1 b−1 , aca−1 c−1 >.
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Proof. For this proof I will include a picture to illustrate the sets and deformation retracts which I will describe here. Consider the two torii as unit squares each with opposite edges associated. To have these torii associated by a circle in each would mean
to intersect these unit squares much like the intersection of two perpendicular planes.
Apply Van Kampen’s thm to this set of intersected squares. Define an open set A which
is the original space but with the opposite edges of one of the square removed, that is
remove a circle from one of the torii. Define a similar set B but with the opposite edges
of the other square removed. Each individual set, A or B, can deformation retract to one
unit square which represents a torus. Thus π1 ( A) ∼
= π1 ( B ) ∼
= Z × Z. The intersection,
A ∩ B, can deformation retract to a line, which represents a circle in the quotient space.
Under inclusions, the normal subgroup we get, I believe, will be Z, the fundamental
group of the circle. By Van Kampen’s thm this gets that the fundamental group of the
(Z×Z)∗(Z×Z)
space is
.
Z
1.2.9
Claim: Mh0 does not retract to C
Proof. Consider the 4g polygon which quotients to become Mh (as described on page 5
of chapter 0). The corresponding polygon for Mh0 would be the same 4g polygon but
with an open circle removed from it’s center. We can deformation retract this radially to
get that π1 ( Mh0 ) is isomorphic to the fundamental group of the image of the perimeter
of this polygon under the quotient map. That is π1 ( Mh0 ) ∼
=< a1 , b1 , . . . , a g , bg >, where
this last group can be thought of as π1 (∨2g S1 ). Note that in this isomorphism the loop
1 −1
of C will correspond to the element a1 b1 a1−1 b1 −1 . . . a g bg a−
g b g i.e. the loop that is the
image of the perimeter of the 4g-gon under the quotient map, call this loop P.
Now assume for a contradiction that there exists a retract from Mh0 → C. This implies
that π1 (C ) ,→ π1 ( Mh0 ). By the previous isomorphism this would imply that π1 ( P) ,→
π1 (∨2g S1 ). From this we can make an exact sequence
0 → π1 ( P) → π1 (∨2g S1 ) → 0
We’ll now apply a functor F which will be abelianization to this exact sequence. This
gets
0 → F (π1 ( P)) → F (π1 (∨2g S1 )) → 0
Since abelianization is a right exact functor this sequence ought to be exact, however,
F (π1 ( P)) cannot inject into F (π1 (∨2g S1 )). Note that F (π1 ( P)) = π1 ( P) since π1 ( P)
corresponds to pi1 (C ) ∼
= Z which is already abelian. Also in the abelianization of
π1 (∨2g S1 ) the element P is one of the elements which gets quotiented out by the commutator. Therefore this cannot be an injection which gives a contradiction.
Since a retract of Mg to C would imply a retract of Mh0 to C by a restriction of the
map, it can’t be that such a retract exists.
Claim: Mg retracts onto C 0
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Proof. Consider the 4g-gon which quotients to Mg as described on page 5 of chapter
0. Note that the circle C 0 corresponds to one edge in the perimeter of this 4g-gon. We
can then iteratively contract corresponding edges in the 4g-gon. For example if the
1 −1
perimeter is labeled a1 b1 a1−1 b1 −1 . . . a g bg a−
g b g then we can first contract the edges a1
and a1−1 to a point and continue with the other edges until we are left with the edge
bg whic, wlog, corresponds to the circle C 0 . Then applying the universal property of
quotient maps, this retraction of the 4g-gon to an edge induces a retraction from Mg to
the circle C 0 .
1.2.10
Proof. I’ll prove this with a series of deformation retracts for which I will attach a set of
drawings. Let X be the space of D2 × I without α ∪ β
1. In a similar manner to the deformation retract of R2 to S1 ∨ S1 we can deformation retract space X to a space made of two hollow cylinders (without tops and
bottoms) linked like horseshoes.
2. We can next deformation retract one of these horseshoe tubes to get the space
which is one tube with a copy of S1 attached at one point
3. Next deformation retract the other tube into another copy of S1 so that the space
we have left is the wedge S1 ∨ S1 .
Finally note that after the series of deformation retracts, the loop γ becomes a path
once around each of the circles in S1 ∨ S1 . Therefore if γ was nullhomotopic in X then it
would be nullhomotopic in the image of the series of deformation retracts, that implies
the path once around each circle in S1 ∨ S1 is nullhomotopic, which is a clear contradiction.
1.2.11
Compute a presentation for the fundamental group of the mapping torus of S1 ∨ S1
Proof. First I will compute the mapping torus of the wedge S1 ∨ S1 following the hint.
Let X = S1 ∨ S1 and consider the CW complex X ∨ S1 . Say the space X ∨ S1 is generated
by the loops a, b, γ, where γ is the loop of the mapping torus, a.k.a. the newly wedge
copy of S1 . We can form the mapping torus by attaching 2 2-cells. The first 2-cell should
be around the path aγ f¯∗ ( a)γ̄ and the second 2-cell should be around the path bγ f¯∗ (b)γ̄.
Now we may apply proposition 1.26 to compute a presentation of the mapping torus.
We know π1 ( X ∨ S1 ) =< a, b, γ >, the fundamental group of the mapping torus is
< a, b, γ| aγ f¯∗ ( a)γ̄, bγ f¯∗ (b)γ̄ >.
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**Note to self : it helped me to think about the case of the mapping torus of a circle
or a disc first**
Compute a presentation of the fundamental group of the mapping torus of S1 × S1
Proof. I got this argument from Alicia (another student in the class). Start with the
wedge of two circles which is created by the loops a, b. Attach a copy of S1 , called
γ which will be a part of the mapping torus. Then attach a 2-cell to create a torus
from the wedge sum. Attach along the path ab āb̄. Next attach 2 2-cells, each along a
respective generator as it is swept through the mapping torus. Similar to before we get
the words aγ f¯∗ ( a)γ̄ and bγ f¯∗ (b)γ̄. By prop 1.26 we can compute a presentation for the
fundamental group as < a, b, γ| ab āb̄, aγ f¯∗ ( a)γ̄, bγ f¯∗ (b)γ̄ >.
1.1.20
Proof. The fudamental group of X can be shown to be the same as ∗n π1 (Cn ) by a similar
argument to what was done to show that the infinite wedge sum is ∗n Z. Take the open
set An to be the circle Cn union a small open nbhd around the origin intersected with X.
So small s.t. the nbhd doesn’t contain any Ci for any i. Then any pairwise intersection
will be path connected and can deformation retract to the origin. Similarly for any triple
intersection. It then follows from the Van Kampen thm that π1 ( X ) = ∗n π1 (Cn ). This is
the same as the fundamental group of the infinite wedge sum because an individual Cn
is homeomorphic to S1 , which implies that the fundamental groups are isomorphic.
I’ll show that the infinite wedge sum of circles is homotopically equivalent to X by
showing that both are CW-complexes with attaching maps which are homotopic (It is
also possible to show that both spaces can be made of deformation retracts of R+ , R
with holes at along the x axis at each positive integer x values).
Consider the set which is the collection of circles of radius n. Call each circle Cn and
the set X0 . For each Cn we can attach this set to the origin by an attaching map f n which
is to simply attach Cn by one point of Cn . There is another attaching map gn which is
to map an arc of Cn to the origin so that the result is a copy of S1 attached to the origin.
Note that both maps f n and gn are homotopic. Then we may combine the maps f n , gn
to get attaching maps F, G respectively for X0 (note here F C = f n , similarly for G).
n
Both F and G will be homotopic simply by doing each homotopy between f n and gn for
all n, simultaneously. Since the attaching maps of a CW complex are homotopic then
the new spaces created will be homotopically equivalent, thus X and the infinite wedge
sum of circles are homotopically equivalent.
The space X is not homeomorphic to ∨∞ S1 because it is not compact in R2 since
it is not bounded, however, the infinite wedge is compact in R2 because it is closed
and bdd. Since homeomorphisms preserve compactness these two spaces cannot be
homeomorphic.
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