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<record version="1" id="5919">
 <title>homotopy invariance</title>
 <name>HomotopyInvariance</name>
 <created>2004-06-15 16:06:17</created>
 <modified>2004-06-15 16:06:17</modified>
 <type>Definition</type>
 <creator id="1001" name="pbruin"/>
 <author id="1001" name="pbruin"/>
 <classification>
	<category scheme="msc" code="55Pxx"/>
 </classification>
 <defines>
	<concept>homotopy invariant</concept>
 </defines>
 <related>
	<object name="HomotopyEquivalence"/>
 </related>
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 <content>Let $\cal F$ be a functor from the category of topological spaces to some category $\cal C$.  Then $\cal F$ is called {\em homotopy invariant} if for any two homotopic maps $f,g\colon X\to Y$ between topological spaces $X$ and $Y$ the morphisms ${\cal F}f$ and ${\cal F}g$ in $\cal C$ induced by $\cal F$ are identical.

Suppose $\cal F$ is a homotopy invariant functor, and $X$ and $Y$ are homotopy equivalent topological spaces.  Then there are continuous maps $f\colon X\to Y$ and $g\colon Y\to X$ such that $g\circ f\simeq{\rm id}_X$ and $f\circ g\simeq{\rm id}_Y$ (i.e. $g\circ f$ and $f\circ g$ are homotopic to the identity maps on $X$ and $Y$, respectively).  Assume that $\cal F$ is a covariant functor.  Then the homotopy invariance of $\cal F$ implies
$$
{\cal F}g\circ{\cal F}f={\cal F}(g\circ f)={\rm id}_{{\cal F}X}
$$
and
$$
{\cal F}f\circ{\cal F}g={\cal F}(f\circ g)={\rm id}_{{\cal F}Y}.
$$
From this we see that ${\cal F}X$ and ${\cal F}Y$ are isomorphic in $\cal C$.  (The same argument clearly holds if $\cal F$ is contravariant instead of covariant.)

An important example of a homotopy invariant functor is the fundamental group $\pi_1$; here $\cal C$ is the category of groups.</content>
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