A monotone function never has saddle points, same is true for an invertible function. Can we conclude that monotonic function is also invertible and vice-versa?
[Math] Does monotonicity of a function imply invertibility? What about vice verse
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For something to be a function, you need one output for each input. You do not need vice versa. Your equation $x^2+y^2=1$ does not define $x$ or $y$ as a function of the other because (for example) if I give you an $x$ there are $0$ or $2$ values for $y$ that satisfy the equation. If you write $y=\sqrt{1-x^2}$ for $-1 \le x \le 1$ you have a valid function because we have restricted the $y$ values to $[0,1]$ and for any $x$ in the domain there is only one $y$ in the range that corresponds. It is still not vice versa because for $y=\frac 35$ you can have either $x=\frac 45$ or $x=-\frac 45$, but that is not a problem.
Maybe a picture will help. Your function does indeed have a minimum at $0$, in fact it is always between $x^4$ and $3 x^4$, but the graph is so wiggly as $x \to 0$ that there are no intervals $(0,\epsilon)$ or $(-\epsilon, 0)$ on which it is monotonic.
Best Answer
Your statement
is false. The function $f:\mathbb R\to \mathbb R$ with $f(x)=x^3$ is (strictly) monotone, has a saddle point at $x=0$, and is invertible with inverse $f^{-1}(y)= y^{1/3}$.
Still, a strictly monotone function $g:\mathbb R\to \mathbb R$ is invertible with its inverse defined everywhere on $g(\mathbb R)$.
Strict monotonicity is required for invertibility. Nondecreasing step functions show why.