# Homotopy extension property in chapter 0 of Hatcher's.

I understand that HEP definition is the following:

$$(X,A)$$ has the homotopy extension property if every pair of maps $$X×\{0\}→Y$$ and $$A×I→Y$$ that agree on $$A×\{0\}$$ can be extended to a map $$X×I→Y$$ .

There is a theorem that says: A pair $$(X,A)$$ has the homotopy extension property if and only if $$X×\{0\}∪A×I$$ is a retract of $$X×I$$ .

For the $$(\Rightarrow)$$ direction, the book says that: Homotopy extension property for $$(X,A)$$ implies that the identity $$X\times \{0\}\cup A\times I\to X\times \{0\}\cup A\times I$$ extends to a map $$X\times I\to X\times \{0\}\cup A\times I$$ so $$X\times \{0\}\cup A\times I$$ is retract of $$X\times I$$.

But I don't understand why that is true. $$(X,A)$$ has the extension property i.e, map on $$A\times I$$ can be extended to $$X\times I$$. But I don't understand what allows the extension in the proof above.

Is the following correct?

Let $$i:X\times \{0\}\cup A\times I\to X\times \{0\}\cup A\times I$$ be the identity map. $$i|_{A\times I}$$ and $$i|_{X\times \{0\}}$$ agree on $$A\times \{0\}$$ so by definition we get an extension $$i_{X\times I}$$ of $$i|_{A\times I}$$. $$i_{X\times I}$$ is the desired retraction map.

• Yes that's what allows it. Commented Mar 5, 2023 at 10:35

Perhaps it is easier to see if we reformulate the HEP as follows:

Each function $$h : X\times \{0\}\cup A\times I \to Y$$ such that $$h \mid_{X\times \{0\}}$$ and $$h \mid_{A\times I}$$ are continuous has a continuous extension $$H : X \times I \to Y$$.

Now take $$Y = X\times \{0\}\cup A\times I$$ and $$h = id$$.

Note that in the above definition we do not require that $$h$$ is continuous, we only require that the restrictions $$h \mid_{X\times \{0\}}$$ and $$h \mid_{A\times I}$$ are. Let us call such a function $$h$$ partially continuous rel. $$(X,A)$$.

By definition, if $$(X,A)$$ has the HEP, then each $$h : X\times \{0\}\cup A\times I \to Y$$ which is partially continuous rel. $$(X,A)$$ must be continuous - otherwise it could not have a continuous extension to $$X \times I$$.

This is a very special feature of pairs with the HEP. For an arbitrary $$(X,A)$$ it is not true that partial continuity rel. $$(X,A)$$ implies continuity. Without assuming that $$(X,A)$$ has the HEP we can only prove that $$h$$ is continuous provided $$A$$ is closed in $$X$$.

For the $$\Leftarrow$$ direction Hatcher gives a simple proof for $$A$$ closed in $$X$$. Indeed, in that case each $$h$$ which is partially continuous rel. $$(X,A)$$ is continuous. Therefore, if $$r : X \times I \to X\times \{0\}\cup A\times I$$ is a retraction, then $$h \circ r$$ is the desired continuous extension of $$h$$.

Observe that Hatcher writes

The hypothesis that $$A$$ is closed can be avoided by a more complicated argument given in the Appendix.