I'm doing some practice problems and am having trouble answering these problems:
Prove or disprove each statement.
(a) If $f : A \rightarrow A$ is one-to-one, then $f$ is onto.
(b) If $A$ is finite and $f : A \rightarrow A$ is one-to-one, then f is onto.
(c) If $f : A \rightarrow A$ is $f$ is onto, then $f$ is one-to-one.
(d) If $A$ is finite and $f : A \rightarrow A$ is $f$ is onto, then $f$ is one-to-one
I know that for a function to be one-to-one there can't be two distinct elements in the domain that map to the same element in the codomain. Also that a function is onto if every element in the codomain has a pre-image in the domain (or in other words every element in the codomain must be mapped to an element in the domain).
Pictorially what I mean is this:
So for my problem: In part a) I believe that just because a function is one-to-one it does not necessarily have to be onto as can be seen from the diagram.
But then I feel like you can use the same logic to disprove all of them, which I'm not sure it correct. (maybe the finite ones are true?)
Any help with these problems would be appreciated.