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A morphism is said to have the homotopy extension property if homotopies out of its domain extend to homotopies out of its codomain.
More in detail, given a category with a concept of path space object and hence of right homotopy, then a morphism $i \colon A \longrightarrow B$ is said to have the right homotopy extension property with respect to an object $X$ if it has the left lifting property against one of the two path endpoint evaluation maps $p_0 \colon Path(X) \longrightarrow X$, i.e. if in every commuting square as below, a diagonal lift exists:
Here the top morphism exhibits a right homotopy between two morphisms $A\to X$, and the diagonal filler is an extension of the top morphism along $i$ and exhibiting a compatible right homotopy between morphisms $B\to X$, whence the terminology.
The concept of homotopy extension property is the Eckmann-Hilton dual of that of (left) homotopy lifting property, where instead one considers the presence of a cylinder object, hence a notion of left homotopy, and lifting in diagrams of the form
In situations where both path space objects as well as cylinder objects exist and are compatible, so that the concepts of left and right homotopy coincide, one may equivalently rephrase right homotopy extension in terms of left homotopy (and left homotopy lifting in terms of right homotopy).
The resulting definition is necessarily less transparent (def. below), but it happens to be more commonly used in the literature. Specifically in the archetypical case that the ambient category is that of topological spaces and cylinders and path space objects are induced from the standard topological interval object, a Hurewicz cofibration (often just called cofibration ) is a continuous function that satisfies the left homotopy extension property with respect to all topological spaces.
(homotopy extension property for left homotopies in topological spaces)
A continuous function $i \colon A \xrightarrow{\;} X$ of topological spaces is said to satisfy the (left) homotopy extension property (HEP) with respect to a space $Y$ if
equivalently:
The map $f$ is sometimes said to be the initial condition of a homotopy extension problem and $\widehat{\eta}$ is called the extension of the homotopy $\eta$ subject to that initial condition.
The map $i$ is called a Hurewicz cofibration if it satisfies this homotopy extension property with respect to all spaces $Y$.
(re-formulation in terms of right homotopies)
In a convenient cartesian closed category $TopSp$ of topological spaces (such as that of compactly generated topological spaces) the product $\dashv$ internal hom-adjunction
allows to pass to adjunct morphisms in Def.
and equivalently express the above diagram of left homotopies as the following (somewhat more transparent) diagram of right homotopies, where $ev_0 \,\colon\, Y^{[0,1]}\to Y$ denotes evaluation at 0 $\gamma \mapsto \gamma(0)$:
In this equivalent formulation, the homotopy extension property is simply the right lifting property against the evaluation map $ev_0 \;\colon\; Y^{[0,1]} \xrightarrow{\;} Y$ out of the path space of $Y$.
(e.g. May 1999, p. 43)
If a map $i\colon A\to X$ has the homotopy extension property with respect to a space $Y$, then for any map $g \colon A\to Z$, the pushout $g_*(i)=i\amalg_A Z :Z\to X\amalg_A Z$ has the homotopy extension property with respect to the space $Y$.
This is a general statement about classes of morphisms defined by a left lifting property, see at injective and projective morphisms – closure properties
We would like to find $\tilde{F}$ to complete the commutative diagram
Consider the external square obtained by composing the horizontal arrows:
By the assumption on $i$, there is a $\tilde{G}$ as in the diagram, such that both triangles commute, i.e. $ev_0\circ\tilde{G}=F\circ i_*(g)$ and $\tilde{G}\circ i = \tilde{F}\circ g$.
If $i:A\to X$ is satisfying the HEP with respect to $Y$ then there is a diagonal in that external square which is some map $\tilde{G}:X\to Y^I$. This map together with $f:Z\to Y^I$, by the universal property of pushout, determines a unique map $\tilde{F}:X\amalg_A Z\to Y^I$ such that $\tilde{F}\circ i_*(g)=\tilde{G}$ and $\tilde{F}\circ g_*(i)=f$. We need to show only that $ev_0\circ\tilde{F}=F$ as $\tilde{F}\circ g_*(i)=f$ holds by the construction of $\tilde{F}$ as stated.
By the definition of $\tilde{G}$ and the commutativity of the original double square diagram, $ev_0\circ \tilde{F}\circ i_*(g)=ev_0\circ\tilde{G}=F\circ i_*(g)$ and $ev_0\circ \tilde{F}\circ g_*(i)=ev_0\circ f=F\circ g_*(i)$. This is almost what we wanted except that we precompose the wanted identity with both maps into the pushout. Thus by the uniqueness part of the universal property of pushout it follows that $ev_0\circ\tilde{F}=F$.
Textbook accounts:
Peter May, Section 6.1 of: A concise course in algebraic topology, University of Chicago Press 1999 (ISBN: 9780226511832, pdf)
Allen Hatcher, p. 14-17 in: Algebraic Topology, Cambridge University Press 2002 (ISBN:9780521795401, webpage)
See also:
The terminology “h-cofibration” is due to: