# How Do Right-Side Homotopies Affect Homotopy Equivalence?

• Ben2
Ben2
Homework Statement
"Show that the composition of homotopy equivalences X-->Y and Y-->Z is a homotopy equivalence X-->Z..."
Relevant Equations
F(x,t) = f_t(x)
"[A] map f: X-->Y is called a \mathbf{homotopy~equivalence} if there is a map g: Y-->X such that fg\cong\mathbb{1} and gf\cong\mathbb{1}," where "cong" means "is homotopic." "The spaces X and Y are said to be \mathbf{homotopy~equivalent}..." Additional definitions are in Hatcher, "Algebraic Topology", of which this is part of Exercise 3(a), p. 18. My difficulty is proving that if f\cong\mathbb{1} and h is another homotopy, then f = f\mathbb{1}h\congh. That is, how do we handle an arbitrary homotopy on the RIGHT? Does this involve "associativity" of homotopies? Thanks!

Apologies for posting! With the conditions given, (hf)(gk) = h(fg)k\congh(\mathbb{1})k = hk\cong\mathbb{1} and (fh)(kg) = f(hk)g\congf\mathbb{1}g = fg\cong\mathbb{1}. This handles transitivity, while the reflexive and symmetric properties are routine. Thanks to everyone who read this!

Ben2 said:
Apologies for posting! With the conditions given, (hf)(gk) = h(fg)k\congh(\mathbb{1})k = hk\cong\mathbb{1} and (fh)(kg) = f(hk)g\congf\mathbb{1}g = fg\cong\mathbb{1}. This handles transitivity, while the reflexive and symmetric properties are routine. Thanks to everyone who read this!

docnet
Ben2 said:
Apologies for posting! With the conditions given, ##(hf)(gk) = h(fg)k\cong h(\mathbb{1})k = hk\cong\mathbb{1}## and## (fh)(kg) = f(hk)g\cong f\mathbb{1}g = fg\cong\mathbb{1}##. This handles transitivity, while the reflexive and symmetric properties are routine. Thanks to everyone who read this!

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