TITLE: Idea behind "reparameterization hiding a corner" in single variable calculus
QUESTION [15 upvotes]: I just solved question #2 on p. 248 from Spivak's Calculus Fourth Edition (2008). Solving it wasn't the issue. I'm trying to understand the idea behind it. This is a screenshot of the question:

I'm trying to understand what is meant by "reparameterizing hides a corner". What does the author mean by "hide" in what sense is it being hidden? For reference, the function that this is being applied to is the following:
$$ f(x) =  \left\{
       \begin{array}\
        x^{2},\ x \geq 0 \\
       -x^{2},\ x \leq 0 \\
      \end{array}  \right.$$
EDIT: Image of fig 21 as requested

REPLY [14 votes]: The other answers cover this nicely. I offer a somewhat different perspective. Notice that the function $h$ is a map from $\mathbb R$ to $\mathbb R$, whereas the parameterization $c$ is a map from $\mathbb R$ to $\mathbb R^2$. This means a priori that the $meaning$ of $h'$ and $c'$ are different. Thus, there is no reason to expect that differentiability of one of them implies differentiablity of the other.
Here is an even worse situation that can occur: take $f(t)=(\cos t,\sin 2t):\ \frac{-\pi}{2}< t< \frac{3 \pi}{2}$. Now, $f$ as defined is injective and differentiable on its domain. The curve (image of $f$) is

If we restrict the curve to $AB$, then the inverse image contains the isolated point $\frac{\pi}{2},$ and so is not open, but $AB$ may be considered as the $graph$ of some $real-valued$ function which is evidently continuous (even differentiable).
The underlying problem is that when we consider continuity and differentiability of $f$ the codomain $\mathbb R^2$ is the space we work in. But when we consider the curve to be the graph of a relation, and restrict it to $AB$ so that is becomes a function, we are changing the codomain to a subset of $f((\frac{-\pi}{2} , \frac{3 \pi}{2})),$ which is quite a different thing. Although $f$ is smooth as a function $(\frac{-\pi}{2}, \frac{3 \pi}{2})\to \mathbb R^2$, $AB$ is not even an open set in the topology induced by $f$, because $f^{-1}(AB)$ is not open!