Question regarding onto and 1-1 functions Let $f:A \to B$ be a map which is both 1-1 and onto. Define the map $g:B \to A$ by the following rule: given y in b, let $g(y)$ equal to the unique element x in A such that $f(x)=y$. Show that g is 1-1 and onto.
I don’t even know how to make sense of this problem. I am completely stuck in understanding the body of the question. 
 A: This problem is essentially asking you to show that bijections (1-1 functions that are onto) have inverses (and that this inverse is also a bijection). Let $f : A \to B$ be a bijection. For every $y \in B$, there exists a unique point $x \in A$ such that $f(x) = y$. This point $x$ exists because $f$ is onto, and is unique because $f$ is 1-1. Define $g(y) := x$, where $x$ is the point with $y = f(x)$. This gives us a function
$$
g : B \to A.
$$
You can verify directly that $g(f(x)) = x$ for all $x \in A$ and $f(g(y)) = y$ for all $y \in B$. Now, you have to show that $g$ is also 1-1 and onto. More precisely, you have to show the following:


*

*$g(y_1) = g(y_2)$ implies $y_1 = y_2$ (i.e. $g$ is 1-1)

*for every $x \in A$, there exists $y \in B$ such that $g(y) = x$ (i.e. $g$ is onto).


To prove these two points,  we will need the following
$$
g(f(x)) = x \quad \text{and} \quad f(g(y)) = y
$$
for all $x \in A$ and $y \in B$. 
We now prove that $g$ is 1-1. Suppose that $g(y_1) = g(y_2)$, applying the function $f$ to both sides of this equality gives:
$$
y_1 = f(g(y_1)) = f(g(y_2)) = y_1.
$$
Hence, $g$ is 1-1. I will leave it to you to verify that $g$ is onto (use that $g(f(x)) = x$ for all $x \in A$).
