Hatcher 0.4

Hatcher, Algebraic Topology, Chapter 0

4. A deformation retraction in the weak sense of a space X to a subspace A is a homotopy f_t:X\to X such that f_0=1_X, f_1(X)\subset A, and f_t(A)\subset A for all t. Show that if X deformation retracts to A in this weak sense, then the inclusion A\hookrightarrow X is a homotopy equivalence.

Proof. For preliminary information regarding the definitions of homotopy equivalences, homotopy inverses, etc., see previous results.

Let X be a topological space, let A\subset X be a subset, and suppose that X deformation retracts weakly onto A by way of the homotopy f_t:X\to X. Let \iota:A\to X denote inclusion. It suffices to show that \iota is a homotopy equivalence, i.e. that there exists a map g:X\to A for which \iota g\simeq 1_X and g\iota\simeq 1_A. Note that f_1(X)\subset A, and so while f_1:X\to X, it’s natural to consider f_1 as a map X\to A. For that reason, the obvious candidate for the map g mentioned above is

g(x)\overset{\text{def}}{=}f_1(x) for all x\in X.

It suffices to show that the this choice of g has the desired properties.

By hypothesis, there is a homotopy f_t:X\to X. Let F:X\times I\to X denote this homotopy (that is, let f_t(x)=F(x,t) for all x\in X,t\in I). It follows, then, that F(x,0)=id_X and that F(x,1)=g(x). In particular, because f_1(X)=g(X)\subset A, \iota\circ g=g for all x\in X. Thus,

F(x,0)=id_X and F(x,1)=f_1(x)=g(x)=\iota\circ g(x) for all x\in X,

and so F is a homotopy connecting f_0=id_X to g=\iota\circ g. Hence, it follows that \iota\circ g\simeq id_X.

Next, define a map G:A\times I\to A by G(a,t)=F(a,t)=f_t(a) for all a\in A, where F is the homotopy above. Then G(a,0)=F(a,0)=f_0(a), i.e. G(\cdot,0)=f_0|_A=id_X|_A=id_A. Similarly, $latex $G(a,1)=F(a,1)=f_1(a)=g(a)=(g\circ\iota)(a)$ for all a\in A. Hence, G is a homotopy connecting g\circ\iota to id_A, whereby it follows that g\circ\iota\simeq id_A. This, combined with the result above, verifies that \iota is a homotopy equivalence with homotopy inverse g=f_1.   \square


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