When $|A|<|B|$ and the function $f: A \to B$ is injective, how do we define the inverse function $g$, with or without the extra elements in $B$?

Question

I need help to understand this elementary part in the set theory. We all have seen the diagram of an injective mapping where few elements of the range set has no pre-image in the domain.

My reasoning gets stuck when I think about how can I define the inverse function $g : B \to A$, where $A, B$ are sets with $|A|=n, |B|=n+2$ and there exists an injective mapping $f: A \to B$, such that each elements of $A$ gets mapped to exactly one element of the set $B$. Clearly set $B$ has extra two elements.

Since $f$ is injective, then $g$ has to be injective.

But my doubts are:-

$a$• when we construct the inverse function $g: B \to A$, then do we include those two un-mapped elements for mapping?

$b$• Is there any difference between these terms $\textrm{not mapping two elements of the domain to the range set}$ and $\textrm{two elements in the domain do not have images in the range set}$ ?

$c$• From the image it is clear that in case of injective mapping there can be elements in the range set who have no pre-image. But can we say that in case of inverse injective mapping some elements in domain (here set $B$) may have no image in the range set ( here set $A$)? If yes, doesn't it contradict the definition of a function?

Also my reasoning in case of constructing the inversion function $g$ in such cases is that the function should look like $g: B \setminus 2 \to A$, to keep the one-one format, but that is not same as $g : B \to A$.

Therefore I request to help me to clear my doubts here. Any response, answer, hints and suggestions are valuable and appreciated.

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Thanks to everyone for suggesting links, notes and the answer.

Answer 1

A function $f: A \to B$ must describe the image $f(a)$ in the codomain $B$ for each $a$ in the domain $A$. This image is unique. It's the image.

This is the forward direction: starting with elements of the domain and thinking about their image. But what about the backwards direction: starting with elements of the codomain and thinking about their preimage?

First of all, this preimage is best defined as a set because it may not be unique or it may not exist: for any $b \in B$, its preimage is $$ f^{-1}(b) = \{ a \in A \mid f(a) = b \}. $$ Despite the suggestive notation, $f^{-1}$ is not necessarily a function in this context.

Here's where two important notions come in.

1. The function is injective (or one-to-one), if for each $b \in B$, its preimage has at most one element. In other words, its preimage is unique.

2. The function is surjective (or onto), if for each $b \in B$, its preimage has at least one element. In other words, its preimage exists.

In order to define an inverse function $g: B \to A$, we need for each $b \in B$ to determine $g(b) \in A$. This element of $A$ must exist and be unique. In other words, the original function $f$ must be surjective and injective. When we have both properties, we call the function bijective (or a one-to-one correspondence). In this situation, it's typical to write $f^{-1} = g$ for the inverse to $f$, satisfying $$ f^{-1}(b) = a \quad\iff\quad f(a) = b. $$