I understand that the inverse of a bijection is a bijection. To proof this you need to formally proof it is injective and surjective. I can prove it is injective and i understand what it is in the following way:
Let $f:A\to B$ be a bijection and let $f^{-1}: B\to A$ be its inverse. To show $f^{-1}$ is a bijection, you must show it is an injection: Let $x_1,x_2 \in B$ such that $f^{-1}(x_1)=f^{-1}(x_2)$. The the inverse we have $x_1= f(f^{-1}(x_2))=x_2$. This shows $f^{-1}$ is injective.
I am having a hard time proving this is surjective formally. I understand this is the definition of a surjection but I dont understand how this applies. Basically I dont understand it.
$$\forall b_i \in B~~ \exists a_j \in A \text{ such that } b_i = f(a_j)$$
Could someone explain the inverse of a bijection, to prove it is a surjection please?
Best Answer
Being a surjection just means you reach all the elements of your target set, here $A$. It is quite easy to show it here: take $x\in A$, then $x=f^{-1}(f(x))$, so $x$ is reached by $f^{-1}$. Therefore $f^{-1}$ is surjective.