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.

  • 2
    $\begingroup$ Yes that's what allows it. $\endgroup$
    – FShrike
    Commented Mar 5, 2023 at 10:35

1 Answer 1


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.


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