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Smooth homotopy

  1. Apr 11, 2012 #1

    jgens

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    Recently I have been working through a text on Differential Topology and have come across the notion of smooth homotopy. Now the textbook (along with every other source I can find on the matter) defines a smooth homotopy of maps [itex]f,g:M \rightarrow N[/itex] as a smooth map [itex]h:M \times [0,1] \rightarrow N[/itex] that satisfies [itex]h(s,0) = f(s)[/itex] and [itex]h(s,1) = g(s)[/itex]. This all makes sense to me except for one thing: The text I am using only defines smooth manifolds without boundary and, unless I am missing something obvious, the space [itex]M \times [0,1][/itex] is not a smooth manifold under this definition. In particular, if [itex]M \times [0,1][/itex] is not a smooth manifold, then our definition of smooth map does not make any sense either. So I am wondering if I am just missing something here, or if there is a genuine problem with this definition.

    Off the top of my head, all of these problems can be alleviated by considering a smooth map [itex]h:M \times (0,1) \rightarrow N[/itex] such that for some [itex]x,y \in (0,1)[/itex] with [itex]x < y[/itex] it follows that [itex]h(s,x) = f(s)[/itex] and [itex]h(s,y) = g(s)[/itex]. This is a bit fussier, but it is immediately clear that [itex]M \times (0,1)[/itex] is a smooth manifold.
     
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  3. Apr 11, 2012 #2

    quasar987

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    Hehe.. I'd say you've got the gist of it.

    Because when we say a map h: M x [0,1] --> n is smooth, we mean that there is an smooth extension of h to some open set, say h: M x (a,b) -->N , where (a,b) contains [0,1].

    More generally, if F is a closed subset of a manifold, a map h:F-->N is smooth if there is an smooth extension h:U-->N where U is open and contains F. In your case, M x [0,1] is a closed subset of the manifold M x R.
     
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