Feb 4, 2021

Notes on the homotopy invariance axiom in (Co)Homology

(Co)Homology theories need to be homotopy invariant, as required by the Eilenberg-Steenrod axioms, that is:

f:X \simeq Y \implies H(f): H(X) \cong H(Y)

In some circumstances, especially when dealing with de Rham cohomology, the homotopy invariance is stated in a different form, as a consequence of Poincarè Lemma, namely:

H(X \times \mathbb{R}) \cong H(X)

via $\pi: X \times \mathbb{R} \to X$ . In this note I review how to reconcile the different formulations of homopoty invariance. The proof is extrapolated from Bott-Tu.

We denote by $I$ the closed unit interval $[0,1]$ (see following remark). We say $f,g: X \to Y$ are homotopic is there is a map $F: X \times I \to Y$ such that $F(\cdot ,0) = f$ and $F(\cdot,1) = g$ . We say two spaces $X$ and $Y$ are homotopy equivalent or homotopic if there exist $f: X \to Y$ , $g: Y \to X$ such that $f \circ g \simeq \mathrm{1}_Y$ and $g \circ f \simeq \mathrm{1}_X$ .

Remark

In the category of smooth manifolds (without boundaries) usually one requires the homotopy to be smooth, and hence usually instead of the closed unit interval, one takes

I

as an open neighbourhood of

[0,1]

\mathbb{R}

itself. In the following then,

I

will denote

[0,1]

or an open neighbourhood of it.

A (co)homology theory is a (co)functor from some category of topological spaces $\mathcal{T}$ ( $\mathcal{T}$ = $\textbf{Top}$ , $\textbf{CW}$ , $\textbf{Diff}$ ,…) to the category of $\mathbb{Z}$ -graded $R$ -modules satisfying Eilenberg-Steenrod axioms. Since the following result uses only functorial properties, we state it for a functor $H:\mathcal{T} \to \mathcal{C}$ , where $\mathcal{C}$ is a general category.

Theorem

For a (co)functor $H:\mathcal{T} \to \mathcal{C}$ the following are equivalent:

For $f,g: X \to Y$ , $f \simeq g \implies H(f) = H(g)$
$f:X \simeq Y \implies H(f): H(X) \cong H(Y)$
$H(X \times I) \cong H(X)$ via $\pi: X \times I \to X$

Proof.

(1. $\implies$ 2.) In fact $X \simeq Y$ means there are $f,g$ such that $f \circ g \simeq \mathrm{1}_Y$ and $g \circ f \simeq \mathrm{1}_X$ . By functoriality $H(f)$ and $H(g)$ are inverses of each other.
(2. $\implies$ 3.) Of course $X \times I \simeq X$ , since the projection on the first factor gives us a retraction.
(3. $\implies$ 1.) Let $F: X \times I \to Y$ be the homotopy $f \simeq g$ . Let $\iota_i: X \to X \times I$ be the inclusion $\iota_i(x) = (x,i)$ . Then $F \circ \iota_0 = f$ and $F \circ \iota_1 = g$ . So we have the commuting diagram:

diagram

Observe that $\iota_i$ is a section of the projection $\pi$ for each $i \in I$ . By applying the functor $H$ , we get $H(\iota_i) = H(\iota_j)$ for $i,j \in I$ , since both are inverses of $H(\pi)$ . This implies

H(F \circ \iota_0) = H(F \circ \iota_1)

Which leads to

H(f) = H(F \circ \iota_0) = H(F \circ \iota_1) = H(g)