Download HOMEWORK 7 Problem 1: Let X be an arbitrary nonempty set

HOMEWORK 7
Problem 1: Let X be an arbitrary nonempty set. Compute the singular homology of X equipped
with the trivial topology {∅, X} and of X equipped with the discrete topology.
Solution: Let Xtriv and Xdisc denote the topological space that consists of X as a set together with
the trivial and the discrete topology, respectively.
For Xtriv observe that Xtriv is homotopy equivalent to a singleton space, so H0 (Xtriv ) = Z whereas
Hi (Xtriv ) = 0 for i > 0. Indeed, fix any element x0 ∈ X and define ρ : X → X, ρ(x) = x0 for every
x ∈ X. Then ρ is homotopic to idX (which proves that ρ is a deformation retraction onto its image)
through the homotopy F : [0, 1] × X → X defined by F (0, x) = x0 and F (t, x) = x for t > 0; this is
continuous as a map of topological spaces because every map Y → Xtriv is continuous.
For Xdisc note that a discrete topological space is homeomorphic to the disjoint union of its points
viewed as singleton spaces. Then Hatcher Proposition 2.6 tells us that
(L
M
x∈X Z (i = 0)
Hi (Xdisc ) =
Hi ({x}) =
{0}
(i > 0).
x∈X
Observe that H0 (Xdisc ) is nothing else but the free abelian group generated by X, i.e. the group of
all finite formal integer linear combinations of elements of X. Said differently, H0 (Xdisc ) = C0 (Xτ )
where Xτ denotes the set X together with an arbitrary topology τ .
Remark: The discussion of Xtriv here is meant to illustrate a comment from lectures—if we drop
the assumption of continuity from the definition of singular simplices, then the resulting “homology
theory” is unable to distinguish any nonempty space from a singleton, in essence because the proof
of homotopy invariance of H• does not use continuity in any meaningful way.
Problem 2: Let U and V be two path-connected open subsets of Rn such that U ∪ V = Rn . Show
that U ∩ V is path-connected.
Solution: This is a straight-up Mayer-Vietoris question. We have a long exact sequence
· · · → H1 (Rn ) → H0 (U ∩ V ) → H0 (U ) ⊕ H0 (V ) → H0 (Rn ) → 0.
Now H1 (Rn ) = 0 since Rn is contractible (or even only since Rn is simply-connected) so H0 (U ∩ V )
injects into H0 (U ) ⊕ H0 (V ) with image equal to the kernel of the surjection φ : H0 (U ) ⊕ H0 (V ) →
H0 (Rn ). Now H0 (U ) ∼
= H0 (V ) ∼
= H0 (Rn ) ∼
= Z since all three spaces are path-connected; conversely
H0 (U ∩ V ) ∼
= Z would imply that U ∩ V is path-connected by Hatcher Proposition 2.6.
It seems intuitively obvious that a surjection Z ⊕ Z → Z must have a “1-dimensional” kernel. In
fact, this would be obvious from the standard rank-nullity theorem if we were using homology with
coefficients in the abelian group G = (k, +), where k is a field—for example k = Q (so this is then
one way of answering the question, assuming that Mayer-Vietoris works with G-coefficients).
Alternatively “rank-nullity for abelian groups” (Google!) shows that H0 (U ∩ V ) has rank 1, and
since H0 (U ∩ V ) is torsion-free by Hatcher Proposition 2.6, H0 (U ∩ V ) ∼
= Z.
Arguably the least pretentious way out is to note that φ(a, b) = a − b if we use augmentation to
identify H0 of each of U, V, Rn with Z, and so then obviously ker(φ) = {(a, a) : a ∈ Z} ∼
= Z.
Note to self: There has to be some way of writing this solution without MV/homology.
Problem 3: Let X = S 1 × S 1 be the torus and Y = S 1 × B 2 the solid torus. Compute the induced
homomorphisms on H1 of the following two continuous maps:
(a) f : X → X, f (z, w) = (z a wb , z c wd ) with a, b, c, d ∈ Z.
(b) The inclusion map i : X → Y of X as the boundary of Y .
2
HOMEWORK 7
Solution: It suffices to compute the induced maps on π1ab , or equivalently on π1 since all fundamental
groups in this question are abelian.
(a) We know that π1 (S 1 × S 1 ) ∼
= Z2 via the isomorphism whose inverse sends (m, n) ∈ Z2 to the
homotopy class of the loop t 7→ (e2πimt , e2πint ). Thus (exactly as in Homework 3) it follows straight
from the definitions that f∗ (m, n) = (am + bn, cm + dn), or in other words f∗ = ( ac db ).
(b) Similarly, what we have proved about fundamental groups is enough to deduce that the map
Z → π1 (S 1 × B 2 ), m 7→ [t 7→ (e2πimt , 1)], is a group isomorphism (using that B 2 is contractible). It
is then once again immediate from the definitions that i∗ (m, n) = m.