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Theorem. Let $X$ and $Y$ be sets with $X$ nonempty. Then (P) there exists an injection $f:X\rightarrow Y$ if and only if (Q) there exists a surjection $g:Y\rightarrow X$.

For the P $\implies$ Q part, I know you can get a surjection $Y\to X$ by mapping $y$ to $x$ if $y=f(x)$ for some $x\in X$ and mapping $y$ to some arbitrary $\alpha\in X$ if $y\in Y\setminus f(X)$. But I don't know about the Q $\implies$ P part.

Could someone give an elementary proof of the theorem?

Martin Sleziak
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ohmygoodness
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3 Answers3

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There is no really elementary proof, since this is in fact independent of the "constructive" part of the usual axioms of set theory.

However, if one has a basic understanding of the axiom of choice, then one can easily construct the injection. The axiom of choice says that if we have a family of non-empty sets then we can choose exactly one element from each set in our family.

Suppose that $g\colon Y\to X$ is a surjection then for every $x\in X$ there is some $y\in Y$ such that $g(y)=x$. I.e., the set $\{y\in Y\mid g(y)=x\}$ is non-empty.

Now consider the family $\Bigg\{\{y\in Y\mid g(y)=x\}\ \Bigg|\ x\in X\Bigg\}$, by the above sentence this is a family of non-empty sets, and using the axiom of choice we can choose exactly one element from every set. Let $y_x$ be the chosen element from $\{y\in Y\mid g(y)=x\}$. Let us see that the function $f(x)=y_x$ is injective.

Suppose that $y_{x}=y_{x'}$, in particular this means that both $y_{x}$ and $y_{x'}$ belong to the same set $\{y\in Y\mid g(y)=x\}$ and this means that $x= g(y_{x}) = g(y_{x'}) =x'$, as wanted.

Some remarks:

The above proof uses the full power of the axiom of choice, we in fact construct an inverse to the injection $g$. However, we are only required to construct an injection from $X$ into $Y$, which need not be an inverse of $g$ -- this is known as The Partition Principle:

If there exists a surjection from $Y$ onto $X$ then there exists an injection from $X$ into $Y$

It is still open whether or not the partition principle implies the axiom of choice, so it might be possible with a bit less than the whole axiom of choice.

However the axiom of choice is definitely needed. Without the axiom of choice it is consistent that there exist two sets $X$ and $Y$ such that $Y$ has both an injection into $X$ and a surjection onto $X$, but there is no injection from $X$ into $Y$.

ACB
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Asaf Karagila
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  • Asaf can you tell me what to look up for more information on the last remark in your answer? When we say 'without the AC..' is that equivalent to saying if we accept the negation of 'AC'? – Prince M May 27 '17 at 04:38
  • Without AC just means without assuming AC. It may hold, or its negation may holds. The issue here is that the negation of AC is just that there is a family which does not admit a choice function. It tells you absolutely nothing constructive about that family. So for the most part, we talk about consistency results. Namely, something is consistent without choice, and on occasion we can say more (e.g. Zorn's lemma fails if AC fails). So what I mean is that it is consistent that the axiom of choice fails, and the Partition Principle fails. So it is not provable just from ZF. – Asaf Karagila May 27 '17 at 12:50
  • https://math.stackexchange.com/questions/4818701/fixing-a-proof-involving-surjective-and-injective-functions – Edward.Lin Dec 02 '23 at 13:29
  • @AsafKaragila What about the $\implies$ direction of the proof when $X = \emptyset$ and $Y \neq \emptyset$? Then $f : X \to Y$ 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 $Y$ to $X$. – Edward.Lin Dec 02 '23 at 13:38
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    @Edward.Lin: Yeah, the equivalence as a whole is for non empty sets. – Asaf Karagila Dec 02 '23 at 13:40
  • Ah, so the precondition is that both $X \neq \emptyset$ and $Y \neq \emptyset$? Great, that's really useful to know and in the textbook I came across the theorem the precondition is not stated. – Edward.Lin Dec 02 '23 at 13:45
  • @Edward.Lin: Trivial conditions are easy to forget. – Asaf Karagila Dec 02 '23 at 13:46
  • @AsafKaragila But it works trivially when both $X = \emptyset = Y$, so the theorem is really that for sets $X, Y$ such that $X = \emptyset = Y$ or $X \neq \emptyset \neq Y$ there is an injection $f : X \to Y$ iff there is a surjection $g : Y \to X$. – Edward.Lin Dec 02 '23 at 13:54
  • @Edward.Lin: Yes. It works for both sets empty. It just doesn't work when one is empty and the other isn't. – Asaf Karagila Dec 02 '23 at 14:18
  • @AsafKaragila But hang on surely "$f: A \to \emptyset$ injective iff $g: \emptyset \to A$ surjective" is true. The ⇒ direction holds because there are no functions $A \to \emptyset$ and the converse is true since there are no surjective functions $\emptyset \to A$? So the condition is for the theorem to hold is that either $A = B = \emptyset$ or $A \neq \emptyset \neq B$ or $A \neq \emptyset = B$. – Edward.Lin Dec 02 '23 at 14:39
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Suppose that $g$ is a surjection from $Y$ to $X$. For every $x$ in $X$, let $Y_x$ be the set of all $y$ such that $g(y)=x$. So $Y_x=g^{-1}(\{x\})$: $Y_x$ is the preimage of $x$. Since $g$ is a surjection, $Y_x$ is non-empty for every $x\in X$.

By the Axiom of Choice, there is a set $Y_c$ such that $Y_c\cap Y_x$ is a $1$-element set for every $x$. Informally, the set $Y_c$ chooses (simultaneously) an element $y_x$ from every $Y_x$.

Define $f(x)$ by $f(x)=y_x$. Then $f$ is an injection from $X$ to $Y$.

Remark: Fairly elementary, I guess, but definitely non-constructive. It can be shown that for general $X$, $Y$, and $g$, the result cannot be proved in ZF$. So we really cannot do better.

André Nicolas
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  • Thank you for your answer. Could you explain why $Y_x=g^{-1}({x})$? How do you know that $g^{-1}$ is well-defined? – ohmygoodness Sep 07 '12 at 18:26
  • @user39561 $Y_x = g^{-1}({x})$ by definition. $g^{-1}({x})$ is always well-defined; however, it may be empty. Using surjectivity, you can show that $g^{-1}({x})$ is not empty because everything has a preimage. Now since $g^{-1}({x})$ is not empty, you now apply the axiom of choice. – William Sep 07 '12 at 18:29
  • So, for example, suppose $Y={a,b}$ and $X={z}$. then there exists a surjection $g:Y\rightarrow X$. Wouldn't $(z,a)\in g^{-1}$ and $(z,b)\in g^{-1}$? Forgive me if I'm being dense. – ohmygoodness Sep 07 '12 at 18:50
  • @user39561: We would have $g^{-1}({z})={a,b}$. Then in the proof we pick an element of ${a,b}$ and send $z$ to that. – André Nicolas Sep 07 '12 at 18:57
  • @user39561: You're quite right that $(z,a),(z,b)\in g^{-1}$. The kicker, here, is that $g^{-1}$ is denoting a relation (set of ordered pairs), and not a function. Both $g$ and $g^{-1}$ are relations, but only $g$ is a function. When we say $g^{-1}({z})$, we are speaking of the image of the set ${z}$ under the relation $g^{-1}$. This isn't the same as saying ${g^{-1}(z)}$--as you pointed out, $g^{-1}$ isn't well-defined. (cont'd) – Cameron Buie Sep 07 '12 at 18:59
  • Put another way, if $h:A\to B$, and $C\subseteq B$, then we define $h^{-1}(C):={a\in A:h(a)\in C}$. Even if $C$ is non-empty, $h^{-1}(C)$ may be empty if $h$ isn't surjective. Even if $C$ is a singleton, $h^{-1}(C)$ needn't be a singleton if $h$ isn't injective. – Cameron Buie Sep 07 '12 at 19:03
  • Ok, got it. Thanks, Cameron and André. – ohmygoodness Sep 07 '12 at 19:13
  • https://math.stackexchange.com/questions/4818701/fixing-a-proof-involving-surjective-and-injective-functions – Edward.Lin Dec 02 '23 at 13:29
  • @AndréNicolas What about the $\implies$ direction of the proof when $X = \emptyset$ and $Y \neq \emptyset$? Then $f : X \to Y$ 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 $Y$ to $X$. – Edward.Lin Dec 02 '23 at 13:39
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This requires the axiom of choice.

Suppose $g : Y \rightarrow X$ is surjective. Then $g^{-1}(x) \neq \emptyset$ for all $x \in X$. By the axiom of choice, there is a choice function $f$ such that for all $x$, $f(x) \in g^{-1}(x)$. $f(x)$ is then the desired injection $X \rightarrow Y$.

Technically, let $\mathcal{A} = \{g^{-1}(x) : x \in X\}$. The choice function is actually a function $\mathcal{A} \rightarrow \bigcup \mathcal{A}$. But I leave it to you to compose it with the appropriate function to get the desired $f$.

William
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