let $A = \{a \in A\}$ and $B = \{b \in B\}$. Let $f$ be a strictly surjective map $f: A \to B$ meaning for every $b$ in $f$'s codomain there must exist some $a$ in $f$'s domain.

$f$ is surjective if and only if it has a right inverse. Why not a left? let $g$ be the inverse map $g: B \to A$ there is no composite function $g\circ f$ that maps $(A \to B) \to A$ Since $f$ is only surjective there can be multiple elements of $A$ that are mapped to the same element in $B$.

Actually it makes sense since $g$ must be a function and its domain would be mapping to multiple co-domain.

What is the reason for a strictly injective function not to have a right inverse?

There cannot be a composite function $f\circ g = (B\to A)\to B$ when $f$ is strictly injective.

if $f$ is strictly injective, there must exist a subset of $A$ that uniquely maps to a subset of $B$ of equal order.

It makes sense that it cannot exist if $f$'s codomain is smaller than $g$'s domain but otherwise it looks fine to me. Can someone explain how this does not make sense?

  • 1
    $\begingroup$ I assume that with "strictly ..." you mean "..., but not bijective"? $\endgroup$ Commented Apr 13, 2016 at 16:09
  • $\begingroup$ yes. $~~~~~~~~~$ $\endgroup$
    – Obliv
    Commented Apr 13, 2016 at 16:10
  • $\begingroup$ The elements of $B$ which are not contained in $f(A)$ are the problem. Any candidate $g$ must map these elements to $A$, but then $f \circ g$ maps them into $f(A)$, so $f \circ g$ cannot be the identity map. If you replace $B$ with $f(A)$ then a right inverse does exist. $\endgroup$
    – user169852
    Commented Apr 13, 2016 at 16:11
  • $\begingroup$ Just a remark on style: "$A=\{a\in A\}$" is literally the same as writing "$A=A$". It would be a bit more elegant to just say "Let $A,B$ be sets$\dots$". $\endgroup$ Commented Apr 13, 2016 at 16:12
  • $\begingroup$ @Bungo I don't think so. Since $f$ is defined as $f: A \to B$ , $B$ is the image of $A$ under $f$ is it not? So $g$ just re-uses this image in its domain. I think the wording is unspecific. $\endgroup$
    – Obliv
    Commented Apr 13, 2016 at 16:20

2 Answers 2


Summarizing the discussion in the comments and chat:

The notation $f : A \to B$ means that $f$ maps each element $x \in A$ to some $y \in B$. In general, not every element of $B$ is "reached" by $f$ in this way. The image of $A$ under $f$, denoted $f(A)$, is the subset of $B$ consisting of the elements of $B$ which are "reached" by $f$.

We always have $f(A) \subset B$, but the containment is proper in general. We have $f(A) = B$ if and only if $f$ is surjective.

Let us suppose that $f$ is not surjective. We will show that in this case, $f$ does not have a right inverse.

Any candidate right inverse $g$ must be a map $g : B \to A$.

Now, since $f$ is not surjective, the set $B \setminus f(A)$ is nonempty. Choose any element $y \in B \setminus f(A)$. Since $y \in B$, it is mapped by $g$ to some element $x \in A$. Then, $f$ maps $x$ to some $z \in f(A)$, by definition of $f(A)$.

Summarizing, we have $y \in B \setminus f(A)$, and $g(y) = x$, and $(f \circ g)(y) = f(g(y)) = z$. Since $z \in f(A)$ and $y \not\in f(A)$, clearly $ z\neq y$. Since $f \circ g$ does not map $y$ to itself, $f \circ g$ cannot be the identity map on $B$. Therefore $g$ is not a right inverse of $f$.

Here is a concrete example. Let $A = [0,1]$ and $B = [2,4]$, and define $f : A \to B$ by $f(x) = x + 2$. Note that $f(A) = [2,3] \subset B$, and the containment is proper, so $f$ is not surjective.

If $g$ is a candidate right inverse for $f$, we must have $g : B \to A$, so in particular, $g$ must map $4$ to some element of $A$, say $g(4) = x \in A$. Then $(f \circ g)(4) = f(g(4)) = f(x) \in [2,3]$ by definition of $f$, and in particular, $(f \circ g)(4)$ cannot be $4$. Thus $f \circ g$ is not the identity map on $B = [2,4]$, so $g$ is not a right inverse of $f$.

Note that we could have defined $f$ the same way except with $B = [2,3]$ instead of $[2,4]$. Then $f$ would be surjective, and $g$ would have a right inverse, namely $g(y) = y - 2$.


Note that $f$ is surjective iff it has a right inverse, and $f$ is injective iff it has a left inverse. In particular, a surjective map that also has a left inverse is both surjective and injective, i.e., bijective. Likewise, an injective function that also has a right inverse is both injective and surjective, i.e., bijective. In other words, a strictly surjective/injective function does not have a left/right inverse.

  • $\begingroup$ Your memory is sharp. Thank you for reciting exactly what my textbook says. $\endgroup$
    – Obliv
    Commented Apr 13, 2016 at 16:21

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