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I cannot prove this rather simple theorem in set theory.

Let $A$ be a finite set. If $f: A \to A$ is injective, then $f$ is also surjective.

My Attempt. Suppose for contradiction that this isnt the case. So there exists a mapping $f: A \to A$ that is not surjective. Then the image of $A$ under $f$ is a proper subset of $A$, i.e., $\text{Im}(f) = C$, where $C \subsetneq A$. Define the mapping $g: C \to A$, where $c \mapsto f^{-1} (\{c\})$. By construction, $g$ is not onto, but since $f$ is injective, it is well-defined, since every $c \in C$ has a unique preimage in $A$. Hence, every $c$ is sent to exactly one $a \in A$. But, if that is the case, $|C| = |A|$, as a function is by definition defined on its entire domain. With this contradiction, we conclude that no such function $f$ can exist.

John P.
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    Does this answer your question? Surjectivity implies injectivity and conversely – Dietrich Burde Feb 28 '20 at 14:18
  • The second of those links is the converse of what I'm trying to prove. The first seems to have access to facts, like the axiom of choice, that I don't. I was hoping someone could look over my proof. – John P. Feb 28 '20 at 14:20
  • @JohnP. To refer to your comment "I was hoping someone could look over my proof". I will try to do so, but first let me try to explain why noone has done this sofar. At this point I can not tell if your proof is correct or not, because I have not thought about it. I have read it, but I have not thought about it. It looks promising so, but that is the problem. Your proof is hard to read due to your choice of making a proof by contradiction. Such proofs are in general confusing to read and you make many claims that you do not elaborate more. With other words your proof is not 'well written'. – Cornman Feb 28 '20 at 16:04
  • I appreciate the feedback. I am working on improving my proof-writing abilities, so any feedback on parts that I should explain more would be greatly appreciated. I will certainly try my best to do so. – John P. Feb 28 '20 at 16:06
  • @JohnP. I do not want to say that my proof below is well written, but I tried my best to make every step made very clear to every reader. I tried to take away the effort to think about my proof below for everyone who reads it, what every proof should attempt and your unfortunatly does not. As I said, I do not want to 'bash' you by any means, it is clear that you put effort into this question and you asked in the 'solution-verification' thread. What I am trying to say is, that it is hard work to read confusing proofs (right or wrong), which is why your request was probably ignored so far. – Cornman Feb 28 '20 at 16:07
  • @JohnP. To talk about your proof: Your first sentence is a little bit odd, even tho it is clear what you mean. "So there exists a mapping $f$ which is not surjective". This is an odd formulation, because the function $f$ you are talking about is not different from the $f$ in the assumption. But it reads like your 'choose' some function $f$ with certain properties. I would not say that it is wrong to write it down like this, but it is (as I said) already a little bit confusing. In the next sentence you do not really come to a point and formulate (more or less) three times the same thing. :) – Cornman Feb 28 '20 at 16:14
  • After all that is not a problem, but again makes it hard to read. Now you define a mapping $g: C\to A$, $c\mapsto f^{-1}({c})$. You claim that this function is by construction not onto (so not surjective). The formulation 'not onto' is again hard to read, because this definition (or formulation) is not very common and I had to look it up. As not a native english speaker I am not as familiar with such expressions, but again not a problem in the proof, makes it just harder to read. :) The problem here is that you do not elaborate more why this function is not surjective by construction. – Cornman Feb 28 '20 at 16:20
  • What I was trying to say was that, because the hypothesis fails, it is not the case that every injective function $f: A \to A$ is also surjective. So there exists an injection that isn't surjective, which I called $f$, and then reasoned about. I don't believe I was invoking the axiom of choice to do so (hopefully it isn't required to do so). – John P. Feb 28 '20 at 16:22
  • @JohnP. I will continue later, I have to leave in a couple minutes. Do not worry to much about the axiom of choice. It is often used without noticing. In this case you do not have to choose a special function (which your formulation implies [at least for me]), because your have in your assumption this injective function $f$ given and now you assume that this function is not surjective. As I said your formulation is not wrong [at least for me], just confusing. I am just trying right now to give you a clue what makes your proof hard to read. I will continue in about two hours. Sorry. – Cornman Feb 28 '20 at 16:26

3 Answers3

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Let $A$ be a finite set and $f:A\to A$ be injective.

We define the (common) notation $f^m(a)=\underbrace{f\circ\dotso\circ f}_{m-times}(a)$ for $m\in\mathbb{N}$.

Since for an injective function $f: X\to Y$ and a function $g: Y\to X$ the function $f\circ g$ is injective, we have that $f^m$ is injective.

Since $A$ is a finite set it exists a $k\in\mathbb{N}$, $k>1$ (the equality $f^1=f$ is trivial) with $f^k=f$ (which means that $f^k(a)=f(a)$ for every $a\in A$). Without loss of generality we can assume that $k\neq 2$. If $k=2$ then $f$ is the identity map, since $f^2(a)=f(f(a))=f(a)$ for every $a\in A$. In that case there is nothing to show.

Why does such $k$ exist? Well, $A$ is finite. That means there is only a finite number of possible functions that can be constructed by concatenating $f$ with itself, since every element can possibly be only mapped at $|A|$ different elements, which means we can at most have $|A|^{|A|}$ possible functions. So if we just concatenate $f$ as often as necessary, we will end up with $f$ again.

As I said it holds that $f^k(a)=f(a)$ for every $a\in A$. So $f(f^{k-1}(a))=f(a)$. But $f$ is injective, which means that $f^{k-1}(a)=a$.

But then we found for every $a\in A$ a preimage of $a$, which is $f^{k-2}(a)$, since $f(f^{k-2}(a))=f^{k-1}(a)=a$. Here we need our assumption without loss of generality, that $k>2$. Else it is not clear that $k-2$ is an element of $\mathbb{N}$ which would make the expression $f^{k-2}$ pointless.

This proof might not be the easiest to find (or even to understand). I tried to make every step clear, which makes the proof look long and complicated.

Cornman
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  • @45465 I do not multiply with $f^{-1}$. I use that $f^{k}=f\circ f^{k-1}$. We have $f^k(a)=f(a)$. Now I write the LHS as $f(f^{k-1}(a))=f(a)$. Now I use that the function is injective, to conclude $f^{k-1}(a)=a$. – Cornman Nov 06 '20 at 19:16
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Your proof is correct, but using a proof by contradiction makes it unnecessarily complicated.

Since $f$ is injective, no two distinct elements of $A$ are mapped to the same element. Hence $|f(A)| = |A|$. As $A$ is finite, $f$ must be surjective.

Guus B
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    Could you explain where we use the assumption of $A$ being finite? Is it because the vertical bars could denote an infinite cardinality, and $|\mathbb{N}| < |\mathcal{\mathbb{R}}|$? – John P. Feb 28 '20 at 14:53
  • That is right. For instance, if $A = (0,2) \subset \mathbb{R}$ and $f$ is given by $x \mapsto x/2$, then the image of $f$ is $(0,1)$, which is not equal to $A$. Still, both cardinalities are the same: $\infty$. – Guus B Feb 28 '20 at 15:00
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Take a copy of the set $A$ and name it $B$. Take an arbitrary element $a_1\in A$ and its image $f(a_1)$ and delete it from the corresponding sets. The new sets $A_1$ and $B_1$ satisfy $|A_1| = |A|-1$ and $|B_1| = |B|-1$. Now take $a_2 \in A_1$ and its image $f(a_2)\in B_1$. Since $f$ is injective $f(a_2) \neq f(a_1)$, hence $f(a_2) \in B_1$. Now delete them from corresponding sets. In the $(2\le) k^{th}$ step take $a_k\in A_{k-1}$. Due to injectivity of the map $f(a_{k}) \notin \{f(a_1),\ldots f(a_{k-1}\} = B\setminus B_{k-1}$, hence $f(a_k) \in B_{k-1}$. Delete them, then leftover sets $A_k$ and $B_k$ satisfy $|A_k| = |A_{k-1}| -1 $ and $|B_k| = |B_{k-1}| -1$. After $|A|$ steps you end up with empty set, i.e. $|A_{|A|}| = \emptyset$. If $B_{|A|}$ was not empty then going back the chain one gets $|B|\geq |A| +1$ which contradicts $|A| = |B|$, since $B$ was a copy of $A$.

Mick
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