2022-12-01

Why is $f\left(n\right)={n}^{3}$ not an onto function?
I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for $f\left(n\right)={n}^{3}$ that it is a one to one function. Then it asked which of the functions from the previous example are onto and $f\left(n\right)={n}^{3}$ was not included in the list of onto functions.
In a later example, a question asked which of these functions is a bijection, the answer included $f\left(x\right)={x}^{3}$
This is confusing because doesn't a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?

Arturo Hogan

Expert

You can never attempt to answer if a function is one-to-one or onto without first knowing the domain and codomain.
For example the function $f:\mathbb{R}\to \mathbb{R},\phantom{\rule{1em}{0ex}}f\left(x\right)={x}^{2}$
is neither one-to-one ($1$ and $-1$ both map to $1$) or onto (nothing maps to e.g. $-1$).
The function
$g:\mathbb{\left[}\mathbb{0}\mathbb{,}\mathrm{\infty }\mathbb{\right)}\to \left[0,\mathrm{\infty }\right),\phantom{\rule{1em}{0ex}}g\left(x\right)={x}^{2}$
is both one-to-one and onto.
We have only changed the domain and codomain, and this changed the properties of one-to-one and onto.

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