Homotopy

Homological algebra初步, 对一些homotopy性质的讨论. Homology函子将homotopic的morphism映射到相同的morphism, 这是以后讨论homotopy需要用到的最根本的性质之一.

以及这个homotopy定义实在是抽象.

Homotopy

Definition 1. A homotopy $h:f\to g$ between two maps $f,g: A\to B$ is a family of morphisms $h_{n}:A_{n}\to B_{n}$ s.t. the following diagram commutes and $f_{n}-g_{n}=d_{n+1}h_{n}+h_{n-1}d_{n}$ for all $n$ .

We write $f\simeq g$ .

We give some basic property of homotopy.

Proposition 1. The homotopy relation is an equivalence.

Proof

Proof.

For tansitivity: If $f\simeq g$ , $g\simeq h$ ,

f_{n}-g_{n}=d_{n+1}s_{n}+s_{n-1}d_{n}

g_{n}-h_{n}=d_{n+1}t_{n}+t_{n-1}d_{n}

So
$f_{n}-g_{n}=d_{n+1}(s_{n}+t_{n})+(s_{n-1}+t_{n-1})d_{n}$ i.e. $f\simeq h$ .

For symmetric: If $f\simeq g$ , $$f_{n}-g_{n}=d_{n+1}h_{n}+h_{n-1}d_{n}\Rightarrow g_{n}-f_{n}=d_{n+1}(-h_{n})+(-h_{n-1})d_{n}$$
For reflectivity: Obvious. ◻

Proposition 2. Let $f\simeq g:A\to B$ , $f'\simeq g':B\to C$ , then $f'f\simeq g'g:A\to C$ .

Proof

Proof. Let $f_{n}-g_{n}=d_{n+1}h_{n}+h_{n-1}d_{n}$ and $f'_{n}-g'_{n}=d_{n+1}h'_{n}+h'_{n-1}d_{n}$ . Then

g'_{n}f_{n}-g'_{n}g_{n}=g'_{n}d_{n+1}h_{n}+g'_{n}h_{n-1}d_{n}=d_{n+1}(g'_{n}h_{n})+(g'_{n}h_{n-1})d_{n}

f'_{n}f_{n}-g'_{n}f_{n}=d_{n+1}h_{n}f_{n}-h_{n-1}d_{n}f_{n}=d_{n+1}(h_{n}f_{n})+(h_{n-1}f_{n}d_{n})

Thus $g'_{n}f_{n}\simeq g'_{n}g_{n}$ and $f'_{n}f_{n}\simeq g'_{n}f_{n}$ , so $f'_{n}f_{n}\simeq g'_{n}g_{n}$ . ◻

Proposition 3. Let $F:\mathrm{Ch}(\mathcal{A})\to \mathrm{Ch}(\mathcal{A})$ be an addictive functor. If $f\simeq g:A\to B$ , then $Ff\simeq Fg:FA\to FB$ .

Proof

Proof. Let $h:f\to g$ . Then we have
$$\begin{aligned}Ff_{n}-Fg_{n}&=F(d_{n+1}h_{n}+h_{n-1}d_{n})\\&= F(d_{n+1})F(h_{n})+F(h_{n-1})F(d_{n})\end{aligned}$$
So $Ff\simeq Fg$ . ◻

Homotopy and homology functor

The homotopic morphisms will induce the same morphism under homology functor $H(-)$ .

Theorem 1. For $f\simeq g:A\to B$ with $h:f\to g$ , we have $H(f)\simeq H(g)$ .

Proof

Proof.
$$\begin{aligned}\mathrm{Im}(f-g)&=(f-g)(\mathrm{Ker}d_{n})\\&=d_{n+1}h_{n}(\mathrm{Ker}d_{n})+h_{n-1}d_{n}(\mathrm{Ker}d_{n})\\&=d_{n+1}h_{n}(\mathrm{Ker}d_{n})\subset\mathrm{Im}d_{n+1}\end{aligned}$$
So $H(f-g)=0$ , i.e. $H(f)=H(g)$ . ◻

There’s an example $f\not\simeq g$ but $H(f)=H(g)$ .

Take the chain complex

Let $f_{0}=id:\mathbb{Z}\to \mathbb{Z}$ and $g_{n}\equiv 0$ , we have $H_{n}(f)=H_{n}(g)=0$ since $H_{0}(A)=Z_{2}$ and $H_{0}(B)=\mathbb{Z}$ .

However, $f\not\simeq g$ , to see this, we tensor $Z_{2}$ on $A$ and $B$ as $\mathbb{Z}$ module.

$H_{0}(A\otimes Z_{2})\cong Z_{2}$ and $H_{0}(B\otimes Z_{2})\cong Z_{2}$ , $H_{0}(f\otimes id)=1$ but $H(g\otimes Z_{2})=0$ . From Proposition 3 above, $f$ and $g$ cannot be homotopic.