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I am reading Analysis on Manifolds by Munkres, in Ch 8 Sec 39. Here is the definition I am looking at:

Defn: Let $A,B$ be open sets in $\mathbb{R}^n, \mathbb{R}^m$, respectively, then the $C^\infty$ maps $g,h:A \to B$ are differentiably homotopic if $\exists$ a $C^\infty$ map $H:A \times [0,1] \to B$ s.t. $\forall x \in A$, we have $H(x,0)=g(x)$ and $H(x,1)=h(x)$, in which case $H$ is a differentiable homotopy between $g$ and $h$.

This is a new definition for me, but when I look at it, it feels all $C^\infty$ maps $g,h:A \to B$ are differentiably homotopic, what came to my mind was to take $$H(x,t)=t \cdot h(x)+(1-t) \cdot g(x)$$ Then as far as I can see $H$ is a differentiable homotopy according to the definition. But I feel I'm missing something, since otherwise why make this definition? Any help is greatly appreciated.

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    This map is not guaranteed to stay in $B$. What your argument correctly shows is that if $B$ is convex, then any two smooth maps into $B$ are differentiably homotopic. – 2017-02-09
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    @QiaochuYuan Thanks a lot for your help. How important is what $B$ is? Can we always take it as large as necessary? For instance, say $f(x)=x$, we can view $f:(0,1) \to (0,1)$ or $f:(0,1) \to \mathbb{R}$ – 2017-02-09
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    So I know in my simple example $f(x)=x$, $B=(0,1)$ happens to be convex, but more generally, can we always just "convex-ify" $B$? – 2017-02-09
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    The term "differentiably homotopic" takes as input the data of the two maps $g, h$, which implicitly includes the data of the choice of $A$ and $B$. Changing $B$ doesn't make sense any more than changing $g$, say: once you have done so you are talking about something other than what you were before. – 2017-02-09

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