If $f:A→B$ is surjective but not injective, it has a left-inverse, but no right-inverse? It would make sense to me if it were the opposite.

This answer says:

If $f:A→B$ is surjective but not injective, it has a left-inverse, but no right-inverse.

But this answer says:

For example let $f: R → [0, ∞)$ denote the squaring map, such that $f(x) = x^2$ for all x in R, and let $g: [0, ∞) → R$ denote the square root map, such that g(x) = √x for all x ≥ 0. Then f(g(x)) = x for all x in $[0, ∞)$; that is, g is a right inverse to f. However, g is not a left inverse to f, since, e.g., $g(f(−1)) = 1 ≠ −1$.

The second answer makes sense to me. But aren't the two statements are saying the negations of each other? So that would make the first answer false?

  • $\begingroup$ As a side note: The inverse only exists if you have some sort of axiom of choice $\endgroup$ – Jürg Merlin Spaak Oct 17 '17 at 6:55

I think there is some confusion on what is left and what is right, stemming from our culture: We like to depict functions as sending things towards the right, and we like to write functions to the left of their argument.

If $f:A\to B$ is surjective, then we can compose it with a $g:B\to A$ to get $B\overset g{\to} A\overset f\to B$ which becomes the identity on $B$. In this case, $g$ is put on the left of $f$ (technically, the existence of $g$ does require the axiom of choice in the general case, but let's not get into that).

However, if we write things algebraically, we denote this function as $f\circ g$ and its value at $b\in B$ as $f(g(b))$, which puts $g$ to the right of $f$.

I believe this is the confusion that gives the two seemingly conflicting answers. For completeness, it is the latter which is the conventional way of writing it, which means that if $f$ is surjective, but not injective, then it does have a right inverse, but not a left inverse.

After reading the first answer, I can also see that he puts the composition in the wrong order (i.e. he writes $f\circ g$ for the function that first uses $f$, then applies $g$ to the result), so this is clearly what has happened.

  • $\begingroup$ Even in this day and age, some people in group theory write the composition of permutations in the wrong order. $\endgroup$ – bof Oct 17 '17 at 7:23
  • $\begingroup$ @bof If I happen to get magically sent back a few centuries and get the oppurtunity to reboot modern mathematics, putting functions to the right of their argument is one of the things I would do (yes, I have a list), which would indeed mean that $f\circ g$ becomes "apply first $f$ then $g$". $\endgroup$ – Arthur Oct 17 '17 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.