It is well-known that the function $f (x) = \dfrac{1}{x}$ is not bijective on the domain of non-negative real numbers (that is, $[0, \infty)$.) , since $f( \cdot)$ may not be well-defined at zero point.
I am curious that what if we admit the convention that $0^{-1} = \infty$ and $\infty^{-1} = 0$.
Then, under this setting, is $f(x)$ bijective on $[0, \infty]$ now?
Could anyone help me out please? Thank you very much in advance!
Yes, if you define $f(0)=\infty$ and $f(\infty )=0$, then your function
$$ f(x)=1/x \text { if $x\in(0,\infty)$ }$$ along with $f(0)=\infty$ and $f(\infty )=0$ will be bijective on $[0,\infty]$
It will map $(0, \infty)$ to $(0,\infty)$ and maps the endpoints to the endpoints.