# Proving that two formulations of the Homotopy Extension Property via diagrams, are equivalent.

Suppose we are working with a collection of topological spaces for which there are a product functor $$F:Set\to Set:X\to X\times I$$ and an exponential functor $$G:Set\to Set: X\to X^I$$ such that $$F\dashv G.$$ In such a collection of spaces, the map $$i : A \to X$$ has the Homotopy Extension property if, for every space $$Y$$, the following extension problem has a unique solution $$\tilde H$$, represented by the dashed line. That is, the square shown below is a pushout.

I am trying to give a careful proof of the fact that there is a solution to the above diagram if and only if there is a solution to the following one:

where $$p_0(g)=g(0)$$ and $$h=GH\circ \eta_A$$ and $$\phi:= G\tilde H\circ \eta_X$$ is the unlabeled morphism along the dotted line in the second diagram.

The second diagram is the one given in an exercise which asked me to prove that the pushout of a cofibration is a cofibration. I was able to do the exercise, but since I am more familiar with the first diagram as regards HEP, I want to see that they are equivalent and I think it's a pure category theory argument. I know we need to use the isomorphism defined by the adjunction, with the naturality square

$$\begin{array}{ccc} \hom( X\times I,Y) & \rightarrow & \hom (X,Y^I) \\ \downarrow & & \downarrow \\ \hom(A\times I,Y) & \rightarrow & \hom(A,Y^I) \end{array}$$

from which we get $$(\tilde H\circ (i\times id))^I\circ \eta_A=i\circ \tilde H^I\circ \eta_X$$ but I can't see how to unravel this. Is my approach correct? How should I continue?

• The first diagram is not required to be a pushout but only a weak pushout. Look up weak colimits. And F and G are functors $\textbf{Top} \rightarrow \textbf{Top}$ not $\textbf{Set} \rightarrow \textbf{Set}$. Jan 11 '20 at 9:13
• Yes, thanks! $\tilde H$ may not be unique. OK. My category theory is rusty----As for the functors, I specified that we were working in suitable topological spaces (like locally compact Hausdorff spaces for example) because in Top, you may not have exponents, right? Jan 12 '20 at 1:05

As Noel Lundström pointed out in his comment, the square in the first diagram is not a pushout diagram because the homotopy extension property does not require that $$\tilde H$$ is unique. If you require uniqueness, then for example $$\{0\} \hookrightarrow I$$ would be not a cofibration. By the way, in the bottom right corner of a pushout square you would not find $$X \times I$$, but the adjunction space $$X \cup_{i_0} A \times I$$ which can be identified with $$X \times \{0\} \cup A \times I$$ if $$A$$ is closed in $$X$$.
You correctly invoke the "adjunction isomorphism" $$\phi : \hom(X \times I,Y) \to \hom(X,Y^I)$$ which is known as the exponential law. There mere existence of some natural bijection $$\phi$$ is not sufficient for our purposes, we need the fact that $$\big(\phi(G)(x)\big)(t) = G(x,t)$$. This implies the essential formula $$p_0 \circ \phi(G) = G \circ i_0$$ because $$(p_0 \circ \phi(G))(x) = p_0(\phi(G)(x)) = \big(\phi(G)(x)\big)(0) = G(x,0) = (G \circ i_0)(x)$$.
The maps $$H$$ in the first diagram and $$h$$ in the second diagram are related by $$\phi(H) = h$$, and so will be the fillers.
1. Let $$\tilde H$$ be a filler in the first diagram. Define $$\tilde h = \phi(\tilde H) : X \to Y^I$$. Since $$\tilde H \circ (i \times id_I) = H$$, we get $$\tilde h \circ i = \phi(\tilde H) \circ i = \phi(\tilde H \circ (i \times id_I)) = \phi(H) = h$$ by naturality. Moreover, $$p_0 \circ \tilde h = p_0 \circ \phi(\tilde H) = \tilde H \circ i_0 = f$$. Hence $$\tilde h$$ is a filler for the second diagram.
2. Let $$\tilde h$$ be a filler in the second diagram. Define $$\tilde H = \phi^{-1}(\tilde h) : X \times I \to Y$$. Then $$\phi(\tilde H \circ (i \times id_I)) = \phi(\tilde H) \circ i = \tilde h \circ i = h = \phi(H)$$ by naturality and we conclude $$\tilde H \circ (i \times id_I) = H$$. Moreover, $$\tilde H \circ i_0 = \phi^{-1}(\tilde h) \circ i_0 = p_0 \circ \phi(\phi^{-1}(\tilde h)) = p_0 \circ \tilde h = f$$. Hence $$\tilde H$$ is a filler for the first diagram.
• Very clear! Thank you. I was missing the forest for the trees. $\phi$ is just the currying iso. Jan 12 '20 at 1:14