# Homotopy Theory and extensions/liftings.

I found the statement: suppose that in the extension problem we have a map f': A -> E homotopic to f, and f' extends. Then it does not follow that f extends. Similarly, if the map g in the lifting problem is homotopic to a map that lifts, it does not follow that g itself lifts. The reader can easily supply counterexamples in both cases.

I don't see what a counterexample would be.

What I do understand that if, in the extension case, if there is another map i:A -> X then we may be able to find some h:X -> E that extends f'. I can't seem to picture how if we have another map H(1) = f:A -> E (as opposed to H(0) = f') where f and f' are homotopic to each other, that one wouldn't have an extension of f also as by the definition(?) of a homotopy there is a continuous mapping between f and f'. But furthermore since they are both mapping to and from the same spaces A and E, why they would be different.

Any thoughts?

Thanks,

Brian

The key concept here is the homotopy extension property; let me quote this from the Wikipedia article of the same name:

Let $X\,\!$ be a topological space, and let $A \subset X$. We say that the pair $(X,A)\,\!$ has the homotopy extension property if, given a homotopy $f_t\colon A \rightarrow Y$ and a map $\tilde{f}_0\colon X \rightarrow Y$ such that $\tilde{f}_0 |_A = f_0$, there exists an ''extension'' of $\tilde{f}_0$ to the homotopy $\tilde{f}_t\colon X \rightarrow Y$ such that $\tilde{f}_t|_A = f_t$.

Let's assume that the pair $(X,A)$ in your question satisfies the homotopy extension property. Let's say that $f:A\to Y$ and $f':A\to Y$ are homotopic via a homotopy $f_t:A\to Y$ and that $f'$ extends to a map $\tilde{f'}:X\to Y$. The homotopy extension property states precisely in this context that there exists an extension $\tilde{f_t}:X\to Y$ of the homotopy $f_t:A\to Y$ such that $\tilde{f_0}=f'$. Of course, in this case $\tilde{f_1}:X\to Y$ is an extension of $f:A\to Y$. Therefore, if $f':A\to Y$ extends, then so does $f:A\to Y$.

Theorem If $(X,A)$ is a CW pair, then $(X,A)$ satisfies the homotopy extension property.

So, any counterexample to your claim cannot be a CW pair.

The entire discussion above also has an analogue which applies to the corresponding question for liftings. In this case, the relevant concept is the homotopy lifting property; e.g., see http://en.wikipedia.org/wiki/Homotopy_lifting_property .

Exercise 1: Show that there is no counterexample that you seek if the pair $(X,A)$ satisfies the homotopy lifting property.

Definition 1 A map $h:E\to B$ is a fibration if it satisfies the homotopy lifting property with respect to every pair $(X,A)$. It is a Serre fibration if it satisfies the homotopy lifting property with respect to every CW pair $(X,A)$.

Definition 2 A map $i:A\to X$ is a cofibration if it satisfies the homotopy extension property with respect to all spaces $Y$.

In other words, you're looking for maps which aren't fibrations and for maps which aren't cofibrations.

I hope this helps!

• Yes, it does help Amitesh. Thanks. But here is a question: What is the difference of an extension and a lifting? Jul 15, 2013 at 7:56
• @Relative0 Brian, an extension is essentially "extending" the domain of a continuous function. E.g., if $f:A\to Y$ is a continuous function and $A\subseteq X$, then an extension of $f$ is a continuous map $g:X\to Y$ such that $g|_{A}=f$. A lifting is essentially "extending or expanding or pushing (to another space)" the range of a continuous function. E.g., if $f:X\to B$ and $p:E\to B$ are continuous functions, then a lifting of $f$ is a continuous map $g:X\to E$ such that $p\circ g=f$. Does that answer your question? It's good, as well, to find examples of your own of these two phenomena ... Jul 15, 2013 at 9:40