# 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 $H_n(X,A)$

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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…

and

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 $H_n (X \cup CA)\xrightarrow{l.e.s} H_n(X \cup CA, CA) \xrightarrow{excision} H_n(X \cup CA - pt, CA - pt) \cong H_n(X,A)$