# Smooth homotopy

1. Apr 11, 2012

### jgens

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 $f,g:M \rightarrow N$ as a smooth map $h:M \times [0,1] \rightarrow N$ that satisfies $h(s,0) = f(s)$ and $h(s,1) = g(s)$. 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 $M \times [0,1]$ is not a smooth manifold under this definition. In particular, if $M \times [0,1]$ 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 $h:M \times (0,1) \rightarrow N$ such that for some $x,y \in (0,1)$ with $x < y$ it follows that $h(s,x) = f(s)$ and $h(s,y) = g(s)$. This is a bit fussier, but it is immediately clear that $M \times (0,1)$ is a smooth manifold.

2. Apr 11, 2012

### quasar987

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.