I'm reviewing some material from the last semester, and in class we gave the following definition:
Def. 1
A pair $(X,A)$ of spaces have the homotopy extension property if for any homotopy $H: A \times I \to T$ such that exists $f: X \to T$ continuous, and such that $H(-,0)= f|_A$, then there exists $\tilde{H}:X \times I \to T$ that extends $H$.
From my understanding, that means that we also have that $\tilde{H}(-,0)=f$; but then we also said that an equivalent definition is
Def. 2
A pair $(X,A)$ of spaces have the homotopy extension property if there exists a continuous map $$r: X \times I \to X \times \{0\} \cup_{A \times \{0\}} A \times I,$$ where $\cup_{A \times \{0\}}$ denotes the pushout.
Given that I have may wrote down the definitions wrong, it seems to me that the two definitions are not equivalent, and in particular the second should be wrong, since for any homotopy $H: A \times I \to T$ such that exists $f: X \to T$ continuous, and such that $H(-,0)= f|_A$, following the second definition we'd not get a map $\tilde{H}:X \times I \to T$ with $\tilde{H}(-,0)=f$, but only a map $\tilde{H}:X \times I \to T$ such that
$$
\tilde{H} \circ i_{X \times \{0\}}\circ i_{A \times \{0\}} = \tilde{H} \circ j_{A \times I}\circ j_{A \times \{0\}},
$$
where $i_{X \times \{0\}}: X \times \{0\} \to X \times I$ is the inclusion, and similarly for $j_{A \times I}: A \times I \to X \times I, j_{A \times \{0\}}: A \times \{0\} \to A \times I, i_{A \times \{0\}}: A \times \{0\} \to X \times \{0\}$.
Could someone please tell me if I'm right and/or clarify the definitions, or give me a more clear definition with the term "extends" properly explained? Thanks!
Best Answer
With the additional assumption that $r$ in Definition 2 needs to be a retraction, both definitions are equivalent:
Given $H: A \times I \to T$ and $f: X \to T$ such that $H(-,0) = f|_A$ allows us to define a map $H' : X \times \{0\} \cup_{A \times \{0\}} A \times I \to T$ which is $f$ on $X \times \{0\}$ and $H$ on $A \times I$. As both maps coincide on $A \times \{0\}$, this is well-defined.
In particular, if there exists a retraction $r : X \times I \to X \times \{0\} \cup_{A \times \{0\}} A \times I $, then we can define $\widetilde{H} = H' \circ r$ and note that $\widetilde{H}$ has the desired property: In fact we have$$\widetilde{H}|_{A\times I} = H' \circ r|_{A \times I} = H'|_{A \times I} = H$$ as well as $$\widetilde{H}|_{X \times \{0\}} = H' \circ r|_{X \times \{0\}} = f$$
This shows that if $(X,A)$ has the homotopy extension property with respect to Definition 2, then it also has the property with respect to Definition 1.
The other direction was already described in the comments: You take $T = X \times \{0\} \cup_{A \times \{0\}} A \times I $ and $H$ and $f$ to be inclusions $A \times I \to T$ and $X \times \{0\} \to T$, respectively. The resulting $\widetilde{H}$ is the desired retraction.