For an injective map between two spaces with the same dimension, does the map need to be linear in order to be bijective?
In other words, if this statement universally true:
For any function, injectivity between same dimension implies bijectivity.
or it is true only for linear functions:
For linear functions, injectivity between same dimension implies bijectivity.
It seems I got some counter example for non-linear functions. So I am primarily interested in if it is true for linear functions.
Best Answer
We can send $\mathbb{R}$ into $\mathbb{R}$ injectively and not bijectively. (It is a bijection to $(-1,1)$).
$x\mapsto\frac{2}{\pi}\arctan(x)$
Now, for linear functions it can also be false. Imagine vector spaces of the same infinite dimension. $S:\ell^2\rightarrow\ell^2$ defined by $S(a_1,a_2,\ldots):=(0,a_1,a_2,\ldots)$ is linear, injective but not bijective.
For linear functions between vector spaces of the same finite dimension then, it is true.
To prove this use the fundamental theorem of linear algebra. See where the function sends a basis of the space.