Confusion on the definition of the inverse function I am currently trying to prove the equivalence of two definitions of enumerability (one involving a surjective function, the other involving an injective one). In doing so, I wanted to consider the inverse of a function, but then it occurred to me that I can't just assume that any function has an inverse. I looked in my lecture notes and got the following definition
Let $f: a \rightarrow b$ be $1$-$1$. Then the inverse function of $f$ is $f^{-1}: Ran(f) \rightarrow \text{Dom}(f)$ such that:  $$\forall \ \ x \in \text{Ran}(f) \ \ \forall \ \ y \in \text{Dom}(f) (f^{-1} = y \leftrightarrow f(y) = x)$$
Where a function $1$-$1$ function is a injective function. i.e.: $f: a \rightarrow b$ is injective ($1$-$1$) iff $$\forall \ \ x,y \in a (f(x) = f(y) \rightarrow x = y)$$
But everywhere else I am reading that only a bijective function has an inverse. Now, unless I am mistaken, being an injective function does not imply being a surjective function, so injective functions are not necessarily bijective. But in this case there is a palpable disagreement between these two definitions, and that affects my proof. So, I wonder: have my notes got it wrong, or have I made a mistake?
Thank you for any help.
 A: Let $f:A \to  B$ be a function with domain $A$ and codomain $B$. This implies every element in $A$ is mapped to a single element in $B$.
Of course, there may be an element in $B$ that is mapped to more than once, and there may be an element in $B$ that is not mapped to at all. If the function is injective, then there are no elements in $B$ that are mapped to more than once, and if the function is surjective, then every element in $B$ is mapped to. If the function is bijective, then it is both injective and surjective. Note that the range $f(A)$ of $f$ is a subset of the codomain $B$ consisting of all elements in $B$ that are mapped to.
Now, the inverse function $f^{-1}: B \to A$ from the codomain $B$ of $f$ to the domain of $A$ of $f$ exists iff $f$ is bijective. Of course, this is because, under $f$, for every element in $B$, we can map backwards to a single element in $A$.
However, if $f$ is injective alone, then we can still define an "inverse" function $f^{-1}: f(A) \to A$ from the range $f(A)$ of $f$ to the domain $A$ of $f$. This is because, under $f$, for every element in the range $f(A)$, we can still map backwards to a single element in $A$, even if there are elements in $B$ that are not mapped to (because the range $f(A)$ only contains those elements in $B$ that are mapped to). But if $f$ is injective alone, then we cannot define an inverse from $B$ to $A$ because there may be an element in $B$ with nothing to map backwards to.
Typically, an inverse function is defined in terms of the domain/codomain of some other function and not the domain/range of said function, so your confusion is understandable.
A: The jective-brothers come in three names, relative to morphisms: injective, surjective and bijective.

*

*The injective brother has always one answer for one question, but there are questions without answers, which it prefers to avoid;

*The surjective brother has oftentimes two or more answers for the same question;

*The bijective brother has always one answer for every question;

In symbolic mathematics, we have that, given domain set as blackboard bold $\mathbb{D}$, image or range set $\mathbb{I}$ and set $\mathbb{F}$ as some super set algebraic structure, we have that it fulfills $\mathbb{D} \to \mathbb{I}, \mathbb{D} \, ? \, \mathbb{F}$. The question mark $?$ is equivalent to:

*

*$\subset$: injective

*$\subseteq$: surjective

*$\equiv$: bijective

