# Reparametrization of curves, inverse of differential $| \frac{dt}{ds}| = | \frac{ds}{dt}|^{-1}$

I'm wondering about the following equality which I found in my Differential Geometry script:

We are given a $\mathcal{C}^l$-curve $\gamma : I \to \mathbb{R}^n$ where $I$ is an interval and $\gamma$ is a function of the variable $t$. If we now reparametrize this curve, i.e. we make $t$ into $t=t(s)$, a diffeomorphism of the variable $s\in J$ ($t: J \to I$, $J$ an interval), then we have by the chain rule

$$\frac{\mathrm{d}\gamma}{\mathrm{d}s} = \frac{\mathrm{d}t}{\mathrm{d}s} \frac{\mathrm{d}\gamma}{\mathrm{d}t}$$

So far so good. But now in my script it is asserted that from this it follows

$$\left|\frac{\mathrm{d}\gamma}{\mathrm{d}s}\right| = \left|\frac{\mathrm{d}t}{\mathrm{d}s}\right| \left|\frac{\mathrm{d}\gamma}{\mathrm{d}t}\right| = \left|\frac{\mathrm{d}s}{\mathrm{d}t}\right|^{-1} \left|\frac{\mathrm{d}\gamma}{\mathrm{d}t}\right|$$

and I don't exactly see why the second equality holds. I think it must have to do with $t$ being a diffeomorphism because for arbitrary functions I'm quite sure this doesn't hold. Can anyone clear this up? Preferably with a proof of $\left| \frac{\mathrm{d}t}{\mathrm{d}s}\right| = \left| \frac{\mathrm{d}s}{\mathrm{d}t}\right|^{-1}$ in the case of diffeomorphisms.

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If we write $t=f(s)$ and $s=g(t)$ where $f,g$ are inverses of one another, then $t=(f\circ g)(t)$. Differentiate both sides wrt $t$ and apply the chain rule to get $$1=f'(g(t)).g'(t)=\frac{dt}{ds}.\frac{ds}{dt},$$ and so $$\left|\frac{dt}{ds}\right|=\left|\frac{ds}{dt}\right|^{-1}.$$