# The difference between inverse function and a function that is invertible?

My discrete mathematics book says:

But I read an answer, https://math.stackexchange.com/a/2415543/390226, said:

["...]There are invertible functions which are not bijective,[..."]

["]A function is invertible if and only if it is injective[."]

So for a function to have a inverse, it must be bijective. But any function that is injective is invertible, as long as such inverse defined on a subset of the codomain of original one, i.e. the image of the original function?

• See Inverse function : "To be invertible a function must be both an injection and a surjection. Such functions are called bijections. The inverse of an injection $f : X → Y$ that is not a bijection, that is, a function that is not a surjection, is only a partial function on $Y$, which means that for some $y ∈ Y, f^{ −1}(y)$ is undefined." This id the case with $\exp$ in the post you have linked. – Mauro ALLEGRANZA Apr 10 '18 at 14:10
• The issue is simply with $A,B$ of $f : A \to B$. If we "build in" domain and co-domain in the def of $f$ we have that the function $\exp : \mathbb R \to \mathbb R$ is not invertible. But there is an old tradition of calling $\log : \mathbb R^+ \to \mathbb R$ its inverse. As per Gentlemen Prefer Blondes: "nobody is perfect". – Mauro ALLEGRANZA Apr 10 '18 at 14:28
• @MauroALLEGRANZA: Thank you so much, now it's clear to me... Because the same book says to define a function, the domain, co-domain, image/range are needed. I didn't know the tradition. – linear_combinatori_probabi Apr 10 '18 at 14:33

It all depends on the co-domain of your function.

When you have a function $$f:A\to B$$ which is one-to-one but not onto $B$, you may restrict your co-domain to a subset of $B'\subset B$ which is the range of $f$.

For example $$f:N \to N$$ defined by $$f(n)=2n$$ is not onto but it is one-to-one.

If we define, $$f^* : N\to 2N$$ with the same definition $f^*(n)=2n$

We have an inverse function, $(f^*)^{-1} (n) = n/2.$

• So when I see "f is invertible", I should connect this idea to "f must be bijective"? If f is not bijective, then I must change its co-domain? – linear_combinatori_probabi Apr 10 '18 at 14:16
• Yes you need a one-to-one correspondence which means bijective. – Mohammad Riazi-Kermani Apr 10 '18 at 14:17

When people said a function is "invertible", they mean it can be made invertible. And the rigorous definition of inverse function of $f$ in my book is: