Hatcher, Algebraic Topology, Chapter 0
4. A deformation retraction in the weak sense of a space to a subspace is a homotopy such that , , and for all . Show that if deformation retracts to in this weak sense, then the inclusion is a homotopy equivalence.
Proof. For preliminary information regarding the definitions of homotopy equivalences, homotopy inverses, etc., see previous results.
Let be a topological space, let be a subset, and suppose that deformation retracts weakly onto by way of the homotopy . Let denote inclusion. It suffices to show that is a homotopy equivalence, i.e. that there exists a map for which and . Note that , and so while , it’s natural to consider as a map . For that reason, the obvious candidate for the map mentioned above is
for all .
It suffices to show that the this choice of has the desired properties.
By hypothesis, there is a homotopy . Let denote this homotopy (that is, let for all ). It follows, then, that and that . In particular, because , for all . Thus,
and for all ,
and so is a homotopy connecting to . Hence, it follows that .
Next, define a map by for all , where is the homotopy above. Then , i.e. . Similarly, $latex $G(a,1)=F(a,1)=f_1(a)=g(a)=(g\circ\iota)(a)$ for all . Hence, is a homotopy connecting to , whereby it follows that . This, combined with the result above, verifies that is a homotopy equivalence with homotopy inverse .