Construct a diffeomorphism that extends the identity map Is there a diffeomorphism  $f:[1,+ \infty) \rightarrow [1,3)$, such that $f$ restricted on $[1,2]$ is the identity map? I think its true, but I don't know how to construct one.

If you want a smooth diffeomorphism you're going to have to get non-analytic functions involved (since the series expansion of $f$ about any point in $[1,2]$ will be the identity function, which is not $f$). Fear not - we can still find an explicit solution.
The first port of call is the famous $$\eta(x) = \cases{ e^{-1/x} & if $x>0$ \\ 0 & if $x \le 0$,}$$ which is a smooth function (check this!) that is increasing on $[0,\infty)$ and converges to $1$. Using this to smoothly interpolate between the identity and the constant 3, we get a smooth function
$$ f(x) = x(1 - \eta(x-2)) + 3\eta(x - 2) = x + (3-x)\eta(x-2).$$
You can check that 
$$f'(x) = 1 - \eta(x-2)\frac{x^2 - 3x + 1}{x^2 - 4x + 4}>0.$$
