Relationship between homotopy pushout and ordinary pushout

algebraic-topologycofibrationshomotopy-theorylimits-colimits

I'm trying to understand the homotopy pushouts and currently looking at the homotopy cofiber.

For two maps $f \colon C \to A$ and $g \colon C \to B$ we defined the homotopy pushout to be the regular pushout after replacing the maps $f$ and $g$ by their cofibrations $C \to M_f$ and $C \to M_g$ where $M_f$ and $M_g$ are the mapping cylinders.

As far as I understand the homtopy cofiber (mapping cone) for a map $f \colon X \to Y$ is the pushout of the diagram $ * \leftarrow X \to M_f$ where the second map is a cofibration. This is the same as the pushout of, for example, $ D^2 \leftarrow X \to M_f$, which is the great thing about homotopy pushouts.

But what is the relationship with these pushouts to the regular pushout of $ * \leftarrow X \to Y$? If there is one in general?

For example if $X = S^1$ and $f$ is the attaching map of the 2-cell of $\mathbb{R} P^2$ to $S^1$ then the regular pushout of the maps $f$ and the inclusion $S^1 \to D^2$ is just $\mathbb{R} P^2$ but what happens when considering the homotopy pushouts?

Thanks in advance 🙂

Best Answer

Given your specific model of homotopy pushout, we can construct a concrete "comparison map" from the homotopy pushout to the usual pushout. Given a span $S=(Y\stackrel{f}{\leftarrow} X \stackrel{g}{\rightarrow} Z)$, denote the pushout by $colim(S)$ and the homotopy pushout by $hocolim(S) := colim(M_f\leftarrow X \rightarrow M_g)$. By the universal property of the pushout for $hocolim(S)$, the compositions $M_f \stackrel{\simeq}{\to} Y \hookrightarrow colim(S)$ and $M_g \stackrel{\simeq}{\to} Z \hookrightarrow colim(S)$ induce a canonical continuous function $hocolim(S) \to colim(S)$. However this map depends quite a bit on the particular expression of $S$.

For example, consider the span $S=(*\leftarrow X \rightarrow *)$, whose pushout is a one-point space. The mapping cylinder of a constant map is the inclusion of the boundary of the cone $X\hookrightarrow CX$, so $$hocolim(S) = colim(CX\leftarrow X \rightarrow CX) = \Sigma X$$ where $\Sigma X$ is the suspension. In this case the comparison map $hocolim(S) \to colim(S)$ is just the constant map $\Sigma X\to *$.

The important point of the above example is that even though there is a homotopy equivalence of diagrams (i.e. homotopy equivalences between the spaces making the appropriate diagrams commute), the pushouts have different homotopy types. In fact one of the main motivations for the homotopy pushout is to have a construction where the homotopy type of the output is invariant under homotopy equivalences of diagrams, and it turns out that taking the pushout of a cofibrant replacement of the diagram does satisfy this invariance. (Here our model of cofibrant replacement is to replace the two target spaces with their mapping cylinders.) Moreover, there is a theorem (see for example Strom's Modern Classical Homotopy Theory Proposition 6.49) which says

If $S = (C\stackrel{f}{\leftarrow} A \stackrel{i}{\hookrightarrow} B)$ is a span such that $i$ is a cobifration, then the comparison map $hocolim(S) \to colim(S)$ is a homotopy equivalence.

In your particular example, the map $S^1 \to D^2$ is a cofibration, so the usual pushout $\mathbb{R}P^2$ actually "is" a homotopy pushout. In fact, the mapping cylinder of $S^1 \to \mathbb{R}P^1$ is the Mobuis strip, so the pushout of the cofibrant replacement is even homeomorphic to $\mathbb{R}P^2$.