# Homotopy pullbacks/relative homotopy groups vs homotopy pushouts/relative homology groups

In Goodwillie's "Calculus I", speaking of a commutative diagram of spaces

$$\begin{array}{c} Y & \rightarrow & Y_1 \\ \downarrow & & \downarrow & \\ Y_2 & \rightarrow & Y_{12} \end{array}$$

there is the following statement

'Cartesian' implies that (for every basepoint in $Y$) the relative homotopy groups of $Y \rightarrow Y_1$ map isomorphically to those of $Y_2 \rightarrow Y_{12}$. Similarly 'co-Cartesian' implies that the relative homology groups of $Y \rightarrow Y_1$, map isomorphically to those of $Y_2 \rightarrow Y_{12}.$

Here Cartesian (resp. co-Cartesian) means that $Y$ ($Y_{12}$) is equivalent via the canonical map to the homotopy limit (homotopy colimit).

What is a more precise sense in which the two parts of the statement about homotopy and homology are similar to each other?

Relative homotopy groups are the homotopy groups of the homotopy fiber. A homotopy pullback square induces an equivalence of the homotopy fibers of two of its parellel maps by the cancellation property of homotopy pullbacks.

Similarly, relative homology groups are the homology groups of the homotopy cofiber. The statement follows by a dual argument.

• Ok, this should be it. Thanks. I first thought whether the passage alluded to some substantial similarity between how homotopy and homology functors behave with respect to homotopy limits and colimits respectively. – Dimitri Chikhladze Sep 24 '16 at 10:17

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories with such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

EDIT: I forgot that the question was asking specifically about relative homotopy and homology. Luckily the correspondence outlined above sends (up to a shift in degree) relative homology to relative homotopy, so everything I said above is still true.

• Can someone explain me the downvote, so I can improve the answer? – Denis Nardin Sep 23 '16 at 18:01