Let $A, B$ be sets and $f : A \rightarrow B$ be a function. If $f$ is injective, show that the function $g : A \rightarrow f(A)$ is bijective where $g(a) = f(a)$ for each $a \in A$.
I can barely make sense of this question, I'm beginning to lose all hope that I will ever be able to understand mathematics at this level.
If $g(a)=g(a')$ then $f(a)=f(a')$ which implies $a=a'$ because $f$ is injective. We have proved that $g$ is injective. If $b \in f(A)$ then there exists $a \in A$ such that $b=f(a)$. But then $b=g(a)$. This proves that $g$ is surjective. Since $g$ is injective and surjective it is bijective.
It makes sense. The image of $f:A\rightarrow B$ is restricted to $f(A)$ and so the restricted function $g:A\rightarrow f(A):x\mapsto f(x)$ is surjective.
Now if $f$ is also injective, then $g$ is injective too. Hence, $g$ is bijective.
This statement is just saying that any application $f:X \to Y$ is surjective if you remove from its range all points that are not the imaxe of some $x \in \mathbb{X}$
Let me maybe try to formulate it more intuitively. A function $f \colon A \rightarrow B$ being injective means that no element in $B$ is „hit“ more than one times. So for every element in $B$ there exists at most one element in $A$ that is sent to the chosen element in $B$. Now surjectivity means that every element in $B$ is „hit“ at least one times. Therefore bijectivity means that every element is „hit“ exactly one time (a 1:1 correspondence). The set $f(A)$ is just the set of all elements that were „hit“ at least one time. Therefore if we restrict the codomain to $f(A)$ every element is „hit“ at least once. As the map $f$ is injective, these elements were also only „hit“ at most one time, which means they were „hit“ exactly one time. Thus we have bijectivity.
