The $n$-th root as a continuous inverse: for a natural $n \ge 1$ the map $x \mapsto x^{n}$ is continuous and strictly increasing on $[0,\infty)$ with image $[0,\infty)$, so its inverse $x \mapsto x^{1/n}$ is continuous and strictly increasing

Let $n \in \mathbb{N}$ with $n \ge 1$ and put $I := [0,\infty)$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and $p : I \to \mathbb{R}$, $p(x) := x^{n}$ (Integer powers $a^m$). Then:

1. $p$ is continuous on $I$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point);
2. $p$ is increasing on $I$ (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences), hence injective (Injection, surjection, bijection);
3. $p[I] = I$;
4. consequently the inverse map $g : I \to I$ of $p$ is continuous on $I$ and increasing, and $g(a) = a^{1/n}$ is the unique nonnegative $n$-th root of $a$ (Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$).

So the $n$-th root function is continuous, and it is obtained from Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g : f[I] \to I$ is continuous and strictly monotone in the same sense as $f$ rather than from a direct $\varepsilon$-$\delta$ estimate.

• Why $n \ge 1$. At $n = 0$ the map $x \mapsto x^{0}$ is constantly $1$ (Integer powers $a^m$), so it is neither injective nor surjective onto $[0,\infty)$, and no inverse exists. Every claim above is stated for $n \ge 1$ and the hypothesis is used in step 1.2.

• Why the domain is $[0,\infty)$ and not $\mathbb{R}$. For even $n$ the map $x \mapsto x^{n}$ is not injective on $\mathbb{R}$, since $(-x)^{n} = x^{n}$, so the continuous inverse theorem does not apply there; restricting to the nonnegative reals is what makes it injective, and it is also where Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ provides the roots.

• What is gained over the root theorem alone. Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ produces the number $a^{1/n}$ for each $a$ separately and says nothing about how it varies with $a$. Claim 4 is the statement that $a \mapsto a^{1/n}$ is a continuous increasing function, and it comes from the structure of the situation rather than from any estimate on roots.