Is It Possible to Invert a Homotopy?

• Ben2
Ben2
Homework Statement
From Hatcher, "Algebraic Topology" p. 3 line 32: "...[T]hree graphs [two circles connected with a line segment, a horizontal figure 8 and an ellipse bisected by a vertical line segment] are all homotopy equivalent since they are deformation retracts of the same space..."
Relevant Equations
##F(x,t) = f_t(x)##, fg\cong\\mathbb{1}##, ##X\congY##, where the author's "congruence" sign has an additional line segment
For F: X x I-->Y, defined by F(x,t) = y, next define G: Y x I-->X by G(y,u) = x. Then for t = u, we have
F[G(y,t),t] = F{G[F(x,t),t]}, which will ideally be ##\mathbb{1}##. Given Hatcher's definitions pp. 2-3, to me it's not clear how to "invert" a homotopy without an inverse function--let alone how to "invert" a deformation retract. The latter seems to be a set of continuous projection maps. Thanks again for all feedback!

Well, if your parameter on ##I## goes from ##0## to ##1##, your inverse would go from ##1## to ##0##, for one.
There may be issues for some contractible spaces, by cardinality alone, i.e., if you contract to a point, you won't be able to invert. Otherwise, " Homotopic" is an equivalence relationship, so that if X is homotopic to Y, then Y is homotopic to X.

Maybe @mathwonk can add something here.

Thanks very much.

WWGD
I agree with WWGD that the point is simply that a deformation retract is a homotopy equivalence, so the issue is that the latter is an equivalence relation (hence symmetric and transitive).
Maybe Hatcher, Cor.0.21, p.16 will be helpful.

Actually, transitivity seems to follow directly from the fact that compositions of homotopic maps are also homotopic., e.g. as on p.2 of these notes:
https://web.northeastern.edu/suciu/U565/U565sp10-homotopy.pdf

Last edited:
WWGD
Thanks to all for the references.

WWGD

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