Quail Entry: Homology Basics

This happened probably closer to mid December, but I’m posting it now

Simplicial and Singular

What a simplicial is roughly captured in the fact that the figure below is a simplicial complex

but if the edges were oriented cyclically then you would not have a simplicial complex (would fail order preserving).  The order preserving bit is probably so that \partial^2 is zero.  Following Hatcher H^\vartriangle_n denotes simplicial homology.

Have isomorphism H_n = H^\vartriangle_n

Natural map H^\vartriangle_n(X,A) \to H_n(X,A)

1. First do case \dim X < \infty, A = \emptyset

use induction, l.e.s for (X^k, X^{k-1})

show directly H^\vartriangle_n(X^k, X^{k-1}) \cong H_n(X^k, X^{k-1}), this finish with 5-lemma

2. For \dim X = \infty show map is inj and surj via

image of chain is compact and hence contained in some X^k; then apply 1.

3. For arb (X,A) use l.e.s for (X,A).

pf provides motivation for relative homology groups H_n(X,A)





Via the above isomorphism, have easy properties

(1)H_n groups are fin. gen.

(2)H_n = 0 for n > \dim X

(easy to see with H^\vartriangle_n)

(3)H_n(X) = H_n(Y) is X,Y homemorphic

In fact, enough X,Y hmtpy equivalent.  This follow from a more general statement about

hmtpy and homology: if f_t \colon X \times I \to Y then f_* := f_{0*} = f_{1*} =: g_*

f_t \colon X \times I \to Y can be used to get p \colon C_n(X) \to C_{n+1}(Y)

s \in C_n(X)

s \times 1 \colon \Delta^n \times I \to X \times I \xrightarrow{f_t} Y becomes

p(s) \colon \cup \Delta^{n+1} \to Y \ \ \ p(s) \in C_{n+1}(Y)

now check f_* - g_* = \partial \circ p + p \circ \partial = 0

simple to see if f_t is hmtpy equivalence get desired iso H_nX \cong H_nY

Relative Homology Groups

Any A \subset X gives rise to s.e.s of complexes

0 \to C_*(A) \to C_*(X) \to \frac{C_*(X)}{C_*(A)} \to 0

and general homological algebras says there is long exact seqence of homology, the third group that appears is labeled H_*(X,A) so get

... \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to ...

(X,A) is a good pair if A is closed and there is open nbd. of A that deformation retracts to A.  This definition is made because of

Good Pair Isomorphism: H_n(X,A) \cong \widetilde{H}_n(X/A)

This is proved via

Excision: Z \subset A \subset X with \bar Z \subset int(A) then (X -Z, A - Z) \to (X,A) gives iso

proved via crazy technical baryocentric subdivision stuff…


l.e.s of a triple (X\supset B \supset A) which arises from the s.e.s of complexes

0 \to \frac{C_*B}{C_*A} \to \frac{C_*X}{C_*A} \to \frac{C_*X}{C_*B} \to 0

for a good pair have X \supset V \subset A and V deformation retracts to A so l.e.s is

H_n(V,A) \to H_n(X,A) \cong H_n(X,V) \to H_{n-1}(V,A)

and middle map is iso because H_n(V,A) \cong H_n(A,A) = 0.

and diagram

where q \colon X \to X/A, the middle q_* doesn’t matter, all that is left to show is that the right q_* is an iso, and this follows essentially because the part where its not an iso is excised.

Disjoint union corollary: (\sqcup X_a, \sqcup x_a) a good pair, get \widetilde{H}_n(\vee X_a) = H_n(\sqcup X_a, \sqcup x_a) = \oplus_a H_n(X_a, x_a) = \oplus_a \widetilde{H}_n(X_a)

And for arbitrary pair (X,A) have a series of isos H_n (X \cup CA)\xrightarrow{l.e.s} H_n(X \cup CA, CA) \xrightarrow{excision} H_n(X \cup CA - pt, CA - pt) \cong H_n(X,A)


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