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Problem statement, as written:

Let $f\colon A\rightarrow B$ be a function. Prove that $f$ is surjective if and only if, for every pair of functions $g,h\colon B\rightarrow C$, if $g\circ f = h\circ f$, then $g=h$.

Question to answer: are there two omissions in the problem statement which need to appear? Namely, $C$ must be a set with more than one element, and $A$ must be non-empty.

I truncated this post because I didn’t receive the answers I was seeking in this post:

Incomplete problem statement in proof regarding functions and compositions

I include it here incase anyone wants to read into my reasoning for my concerns above.

Cyrus
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1 Answers1

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Question to answer: are there two omissions in the problem statement which need to appear? Namely, $C$ must be a set with more than one element, and $A$ must be non-empty.

The implication that $C \ne \emptyset$ isn't particularly necessary. Why? Suppose $C = \emptyset$. But then you don't have functions $g,h : B \to C$ that your question supposes, if $B \ne \emptyset$. That is, there are no functions from nonempty sets to the empty set. And we can't have a function $f : A \to B$ if $B = \emptyset$ either, for the same reason, if $A$ is nonempty.

Thus, if $C$ is empty, so is $B$, and thus so is $A$ (and yes, there does exist a function $\emptyset \to \emptyset$).

Okay, fine. Does the proposition still hold? Yes. We note: any function $f : \emptyset \to \emptyset$ is unique. Moreover, it is trivially a bijection (the cardinality of its domain and codomain are equal). Thus the proposition absolutely holds, as $f \circ f = f \circ f$ holds trivially, and gives $f=f$ whenever $f$ is invertible (which it is here).

Thus, specifying the details for whether $C$ or other sets are empty isn't necessary. They can be accounted for, and while they require some vacuous reasoning, everything still holds. The most they would affect is the existence of the functions $f,g,h$ in question and give restrictions on the sets $A,B,C$ in question.

Some further reading:

  • Existence of function $\emptyset \to A$ (which can be empty or not): link
  • Uniqueness of that function: link
  • Nonexistence of function $A \to \emptyset$ for $A$ nonempty: link
PrincessEev
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  • Everything you said is correct (and I regret to say I was priorly aware of).. but I think you may have slightly misread my question . I wrote that it’s necessary that $C$ be a set with MORE than one element. – Cyrus May 07 '20 at 05:47
  • And I established that it can work even if $C$ is an empty set. $C$ could be a singleton too, if you prefer, but my point is that cardinality doesn't matter here. – PrincessEev May 07 '20 at 05:48
  • My proof requires that $C$ contains at least two elements. Have you devised a proof with the explicit possibility that $C$ be a singleton? Asking respectfully – Cyrus May 07 '20 at 05:51
  • There is also the possibility that $A$ is empty, but not $B$. This case concerns me because, if so, $f$ cannot be surjective by hypothesis, nor can it be proven to be. Am I correct? I understand it would be quite cumbersome (and distracting) to write something like “$A$ is non-empty, unless $B$ is empty.” I’d just like to confirm my reasoning is correct. – Cyrus May 07 '20 at 06:02