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I'm trying to prove that there exists an injective function $f: A \to B$ if and only if there exists a surjective function $g : B \to A$. I'm fine with the [⇐] direction (which requires the Axiom of Choice), but stuck with part of the [⇒] direction. The theorem is discussed here There exists an injection from $X$ to $Y$ if and only if there exists a surjection from $Y$ to $X$. but the answers there do not touch upon the issue I have. Here is the start of my proof:

[⇒]

Let $f: A \to B$ be an injection. We need to prove there exists a surjective function $g: B \to A$.

Cases:

  1. $A \neq \emptyset = B$.
  2. $A = \emptyset = B$.
  3. $A \neq \emptyset \neq B$.
  4. $A = \emptyset \neq B$.
  • In case 1 the implication follows because the antecedent is false (there cannot be a function from a non-empty set into the empty set).
  • In Case(2), $f^T$ is trivially a surjective function.
  • For Case 3 we take some arbitrary but fixed element $z \in A$ and then set

$$g =_{df} f^T \cup (B \setminus \operatorname{range}(f) \times \{z\})$$

If $\operatorname{range}(f) = B$ then $g = f^T \cup (B \setminus \operatorname{range}(f) \times \{z\}) = f^T \cup (\emptyset \times \{z\}) = f^T$. And $g$ is a surjective function.

  • But now consider Case 4. In this case $f : \emptyset \to B$ is clearly an injective function. But since there can be no function from a non-empty set into the empty set, there cannot be a function from $B$ to $A$, so there cannot be a surjective function.

So how can I find a surjective function in Case 4?

Asaf Karagila
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Edward.Lin
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  • How do you define a function from an empty set? – Yathi Dec 02 '23 at 13:01
  • For any set $A$ there is a function from the empty set to $A$, simply because trivially for every member $x$ of the empty set there is a $y \in A$ such that $(x, y) \in \emptyset \times A$. Likewise the condition of right uniqueness is trivially met (for $(x, y) \in \emptyset \times A$ and $(x, y') \in \emptyset \times A$ trivially imply $y = y'$). – Edward.Lin Dec 02 '23 at 13:06
  • When the set is empty, how can you say 'for every member $x$ of empty set'? – Yathi Dec 02 '23 at 13:14
  • Or is it that for any subset $S$ of $A$, $\emptyset \times S$, is a function from $\emptyset$ to $A$? – Yathi Dec 02 '23 at 13:15
  • It's trivially true that for any set there is a function from the empty set to because the antecedent of the conditional is always false in the definition of function. And a conditional is true when the antecedent is false. So $\forall x (x \in \emptyset \Rightarrow \exists y(y \in A \land (x, y) \in \emptyset \times A ))$ is always true. Likewise for $\forall x \forall y, y' (x \in \emptyset \land y,y' \in A \land (x, y) \in \emptyset \times A \land (x, y') \in \emptyset \times A \Rightarrow y = y' )$. – Edward.Lin Dec 02 '23 at 13:18
  • @YathirajSharma you can always truthfully say "for every x in the empty set, blah-blah", it's vacuously true. That's how the conditional (if-then) works. Look it up. – BrianO Dec 02 '23 at 19:51

1 Answers1

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You can’t. If $B$ is non empty, there is no surjective function $g:B\to\emptyset$ because there is no function at all. As you already stated in (1).

Antonio
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  • Ok but then how can the theorem be correct? According to the theorem, for any sets $X, Y$, there is an injective function from $X$ to $Y$ if and only if there is a surjective function from $Y$ to $X$. It is the theorem discussed here: https://math.stackexchange.com/questions/192460/there-exists-an-injection-from-x-to-y-if-and-only-if-there-exists-a-surjecti – Edward.Lin Dec 02 '23 at 13:33
  • @Edward.Lin that proof is indeed assuming non-emptyness. – Julián Dec 02 '23 at 13:47
  • @Julián Non-emptiness of both $A$ and $B$, or just non-emptiness of one of them? – Edward.Lin Dec 02 '23 at 13:48
  • @Edward.Lin Because it is vacously true. For every $f:\emptyset\to B$ (Injective) there exists $g:B\to\emptyset$ (Surjective). – Antonio Dec 02 '23 at 13:49
  • @Antonio That's not right. There is no vacuous truth going from $f: \emptyset \to B$ to $g : B \to \emptyset$. $g$ in that case cannot be a function. So the theorem requires as a precondition that either $A = \emptyset = B$ or $A \neq \emptyset \neq B$, and this is what has been confirmed to me in the comments here https://math.stackexchange.com/questions/192460/there-exists-an-injection-from-x-to-y-if-and-only-if-there-exists-a-surjecti – Edward.Lin Dec 02 '23 at 13:57
  • I agree. So your claim in (3) was false. – Antonio Dec 02 '23 at 14:01
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    @Edward.Lin (For $A$ and $B$ non-empty.) "$f:∅ → B$ injective implies $g:B → ∅ $ surjective" is definitely a problem by the answer. But "$f:A → ∅ $ injective iff $g:∅ → A$ surjective" is fine since $\Rightarrow$ is true because there are no functions $A → ∅ $ and the converse is also true since there are no surjective functions $∅ → A$. Having said this, I find this just a coincidence. Morally you just want to think of both $A$ and $B$ as non-empty. – Julián Dec 02 '23 at 14:13
  • @Antonio I wasn't making a claim in listing 3., I was discussing that case. in that case (when $A = \emptyset \neq B$) the theorem does not apply. – Edward.Lin Dec 02 '23 at 14:23
  • @Edward.Lin Ok. I intended that the statement in OP doesn’t apply to case (3) $A=\emptyset\neq B$. And my answer to the question in OP was correct, you can’t find a surjective etc. etc. – Antonio Dec 02 '23 at 14:57
  • @Antonio So we were agreeing :) – Edward.Lin Dec 02 '23 at 15:07
  • Yes, we were. (: – Antonio Dec 02 '23 at 15:27