Does a function need to be either surjective, injective, or bijective to be invertible?

Question

I believe a function has to be bijective to be invertible, in fact Wikipedia says that an invertible function is another name for a bijection. https://en.wikipedia.org/wiki/Bijection

But in answers to the question Is a bijective function always invertible?, it is stated that a function only has to be injective to be invertible.

Answer 1

That depends on subtleties of your definition of inverse function. If you expect a function $f\,:\, A \to B$ to have an inverse $f^{-1} \,:\, B \to A$, then $f$ needs to be surjective. If it is not, there are some $b \in B$ that aren't reached by $f$ at all, so how would you define $f^{-1}$ for such elements of $B$?

However, any injective $f\,:\, A \to B$ can be made surjective by simply restricting $B$, i.e. by setting $\widetilde{B} = \{f(a)\,:\, a\in A\}$ and redefining $f$ as $f \,:\, A \to \widetilde{B}$. That redefined $f$ then has an inverse $f^{-1} \,:\, \widetilde{B} \to A$.

Note, however, that while this makes the concatenation $f^{-1} \circ f$ the identity funtion on $A$, it does not make $f \circ f^{-1}$ the identity function on $B$, only on $\widetilde{B} \subsetneq B$.

Thus, if $f \,:\, A \to B $ is injective but not surjective, it has a left-inverse, because we can find an $f^{-1}$ as above with the property that $f^{-1} \circ f$ is the identity on $A$. But it does not have a right-inverse, because we won't find a $f^{-1}$ such that $f \circ f^{-1}$ is the identity on $B$ - the best we can achieve is an identity on $\widetilde{B}$.

Similarly, if $f \,:\, A \to B$ is surjective but not injective, it has a right-inverse, but no left-inverse.

Answer 2

A function has an inverse if and only if it is both surjective and injective. (You can say "bijective" to mean "surjective and injective".)

Khan Academy has a nice video proving this.

edit: originally linked the wrong video.

Answer 3

Hint: if function $ f : A \rightarrow B $ was not surjective, how would we define $ f^{-1} : B \rightarrow A $ for an element that was not in the image of $ f $?

Answer 4

An invertible function shall be both injective and surjective, i.e Bijective! where every elemenet in the final set shall have one and only one anticident in the initial set so that the inverse function can exist!

Answer 5

If your function $f: X \to Y$ is injective but not necessarily surjective, you can say it has an inverse function defined on the image $f(X)$, but not on all of $Y$. By assigning arbitrary values on $Y \backslash f(X)$, you get a left inverse for your function.

Answer 6

This depends on the definition of inverse, which varies in the literature across different areas of mathematics. In probability theory, the inverse of $g:\mathcal{X} \to \mathcal{Y}$ is often defined even though $g$ is not bijective. (The argument below requires no formal knowledge of probability).

Suppose we have a random variable $X$ (a function) defined on on some sample space $\Omega$ such that $X : \Omega \to \mathcal{X}$. Furthermore, define the function $g:\mathcal{X} \to \mathcal{Y}$ for some set $\mathcal{Y}$, not necessarily bijective.

If we define the function $Y := g \circ X$, then it is clear that $Y: \Omega \to \mathcal{Y}$. The inverse function $g^{-1}: \mathcal{Y} \to 2^{\mathcal{X}}$ is defined such that $g^{-1}(y) = \{x\in\mathcal{X} : g(x) = y \}$, which can be the empty set if $y \in \mathcal{Y}$ is not mapped under $g$.

This definition is extremely useful for finding distributions of transformations of random variables.