I'm reading through Dummit and Foote, and they have the following statements about functions and inverses.

Given a map $f: A\to B$ :

  • The map $f$ is injective if and only if $f$ has a left inverse.
  • The map $f$ is surjective if and only if $f$ has a right inverse.
  • The map $f$ is a bijection if and only if there exists $g: B\to A$ such that $f\circ g$ is the identity map on $B$ and $g\circ f$ is the identity map on $A$.

Where $f$ has a left inverse if there exists a function $g: B\to A$ such that $g\circ f$ is the identity on $A$ and $f$ has a right inverse if there exists a function $h: B\to A$ such that $f\circ h$ is the identity on $B$.

The relation between inverses and injectiveness/surjectiveness/bijectiveness make sense to me intuitively. But in particular, the bijection case requires that the right and left inverse be the same. Does this mean that there exists a function $f$ that is not a bijection which has distinct left and right inverses? Or would such a function require a quantifier of the domain of its inverses? The only thing I can tell is that where the domain of the left and right inverses overlap, they must be equal. If the overlap of the domains is $B'$, then $f$ must be injective on $B'$ (because it has a left inverse in that domain), and $f$ must also be surjective (because it has a right inverse in that domain). Therefore, $f$ is bijective on $B'$ and $\forall a\in B', g(a)=h(a)$. What else can be concluded?


1 Answer 1


If $f:A\to B$, then a left inverse $l:B'\to A$ for some $B'\subseteq B$ has to satisfy $l\circ f=\operatorname{id}_A$, which means it has to be defined on the image of $f$ (so $B'\supseteq f(A)$). If $f$ has a right inverse $r:B\to A'$ for some $A'\subseteq A$ such that $f\circ r=\operatorname{id}_B$, then in particular, $f$ is surjective on $B$, which forces that the domain of the left inverse has to be $B'=B$. Moreover, the existence of the left inverse will force $A'=A$: as the left inverse makes $f$ injective, it follows that any $a\in A$ is the unique element that realises $b := f(a)\in B$, so $r(b)=a$ is forced to ensure $f(r(b))=b$; in other words, any $a\in A$ lies in the image of $r$, forcing $A'=A$.

Finally, if a function $f:A\to B$ has a left inverse $l:B\to A$ and a right inverse $r:B\to A$, then in fact $r=l$ and $f$ is bijective. Indeed, this follows by the computation $$ l = l\circ\operatorname{id}_B = l\circ(f\circ r) = (l\circ f)\circ r = \operatorname{id}_A\circ r = r $$ showing that a left inverse and right inverse must always coincide when they both exist.

Conclusion: If $f:A\to B$ has a left inverse $l:B'\to A$ for some $B'\subseteq B$ and a right inverse $r:B\to A'$ for some $A'\subseteq A$, then $B'=B$, $A'=A$, $r=l$, and $f$ is bijective.

Remark. The domain of a left inverse technically only has to be defined on the image of $f$, but we can extend it to an arbitrary superset of this, and it would be free to act however it likes on this set, and a right inverse would be unable to control this. For example, let $f:\{1,2\}\to\{1,2\}$ be the identity, then we can define a left inverse $l:\{1,2,\dots,n\}\to\{1,2\}$ that sends all $k\geq2$ to $2$. However, we can't do much in the case of a right inverse: the image of $r$ has to fit in the domain of $f$ or else the composite $f\circ r$ will not be well-defined. Therefore, perhaps a more correct formulation of the conclusion is:

Conclusion (revised): If $f:A\to B$ has a left inverse $l:B'\to A$ for $B'\supset f(A)$ and has a right inverse $r:B\to A'$ for $A'\subseteq A$, then

  • $A'=A$
  • $f(A)=B$
  • $f$ is bijective with inverse $l|_B=r:B\to A$

Remark. I should mention that left and right inverses usually have their co/domains set by definition based on the co/domain of $f:A\to B$ (namely, they're defined to be maps $B\to A$) because being more flexible doesn't provide any additional benefit in the context of inverses of $f:A\to B$. The goal of a left inverse is to provide a rule for converting elements $b\in B$ into elements $l(b)\in A$ in a way that "undoes" $f$; that is, $f(a)$ gets sent back to $a$. If we look beyond elements of $B$ when defining $l$, this is beyond the scope of $f$ and what we do here is arguably irrelevant.


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