Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From wikipedia I obtain the following definition of an injective function :

Let $f$ be a function whose domain is a set $A$. The function $f$ is injective if for all $a$ and $b$ in $A$, if $f(a) = f(b)$, then $a = b$; that is, $f(a) = f(b)$ implies $a = b$.

From this I conclude that a function $f$ is injective if the below statement is true for all $a,b \in A$:

$$f(a)=f(b) \implies a=b$$

My question is: Can I re-formulate the above statement as $f(a)=f(b) \iff a=b$ ?

share|cite|improve this question
$a = b \implies f(a) = f(b)$ is already part of the definition of a function since a function assigns exactly one output to each input. – user17762 Oct 26 '11 at 21:10
Thus: Yes, your reformulation will also work. – Henning Makholm Oct 26 '11 at 21:12
Sivaram: that's not really part of the definition of functions... If $a$ is equal to $b$, then how could possibly $f(a)$ be not equal to $f(b)$? That's waaaay below, in the rules of manipulation of equality. (notice that for all this to make sense one has to define what $f(a)$ is, but even with one-to-many 'functions' it is true that $a=b\implies f(a)=f(b)$!) – Mariano Suárez-Alvarez Oct 26 '11 at 21:15
Part of the definition of a function is $(a,b) \in f$ and $(a,c) \in f$ then $b=c$. This is exactly the condition $a=b$ implies $f(a)=f(b)$. – Bill Cook Oct 26 '11 at 21:27
@Arturo: but that is a different statement alltogether! :D The meaning of the notation $f(a)$ cannot depend on whether the relation $f$ is or not a function, if the implication $a=b\implies f(a)=f(b)$ is to have sense as part of the definition of what it means for a relation to be a function. I claim that for any definition of what the notation $f(a)$ means for a relation $f$ and an element $a$, the implication will hold for all relations. – Mariano Suárez-Alvarez Oct 26 '11 at 21:31
up vote 4 down vote accepted

Yes, but the implication $a=b\Rightarrow f(a)=f(b)$ holds because you have a function. So in a sense it is redundant.

That is, "$f$ is a function" implies "$a=b\Rightarrow f(a)=f(b)$". So $$\begin{align*} f\text{ is an injective function} &\text{is equivalent to } f\text{ is a function and f is injective}\\ &\text{is equivalent to } f\text{ is a function and } f(a)=f(b)\Rightarrow a=b\\ &\text{implies }\Bigl( (a=b\Rightarrow f(a)=f(b))\text{ and }(f(a)=f(b)\Rightarrow a=b)\Bigr)\\ &\text{is equivalent to } f(a)=f(b)\Leftrightarrow a=b. \end{align*}$$ Conversely, since "$f$ is a function and $a=b \Rightarrow f(a)=f(b)$" is equivalent to "$f$ is a function", you also have the implication going the other way provided you know that $f$ is a function.

share|cite|improve this answer


Let us see why:

$f$ is a function if:

  1. $f\subseteq A\times B$ for some sets $A, B$. That is the elements of $f$ are ordered pairs.
  2. For all $a\in A$ there exists $b\in B$ such that $(a,b)\in f$.
  3. For every $a\in A$ if $b,c\in B$ such that $(a,b)\in f$ and $(a,c)\in f$ then $b=c$. We then denote this unique $b$ as $f(a)$.

This means that if $x=y$ then $f(x)=f(y)$ by the definition of a function.

Therefore for injective functions, it is enough to require $f(x)=f(y)\Rightarrow x=y$, since by the definition of $f$ as a function we have the other direction.

share|cite|improve this answer

Yes, you can. You can formally prove that if a=b, then f(a)=f(b), where f denotes any unary predicate, as follows:

1 |a=b hypothesis
2 |f(a)=f(a) equality (identity) introduction
3 |f(a)=f(b) equality elimination 1, 2, or replacing "a" on the right by "b"
4 If a=b, then f(a)=f(b) 1-3 conditional introduction

So, if f also denotes a function, then a=b implies f(a)=f(b). Thus, f(a)=f(b)⟹a=b can get reformulated as f(a)=f(b)⟺a=b

share|cite|improve this answer

Yes. $a=b$ $\Longrightarrow$ $f(a)=f(b)$ means the function $f$ is well-defined.

By definition every function is well-defined.

Example: $f:\mathbb{Q} \to \mathbb{Z}$ "defined" by $f(a/b)=a$ (the numerator "function") is not well-defined since $4/2=6/3$ but $f(4/2)=4$ and $f(6/3)=6$. The same input gave two different outputs. Thus $f$ is not well-defined. It's not a function!

share|cite|improve this answer
The sentence "$f$ is well-defined" makes no sense: if something is not well-defined, it cannot even be the subject of a sentence. What one really means is that "the relation $f$ which we have defined is a function". – Mariano Suárez-Alvarez Oct 26 '11 at 21:18
And notice that if $f\subseteq X\times Y$ is a any relation whatsoever from $X$ to $Y$ and you define, as usual, for each $x\in X$ that $f(x)=\{y\in I:(x,y)\in R\}$, then it is also true that $x=x'\implies f(x)=f(x')$... so that implication is not characteristic of functions, really. – Mariano Suárez-Alvarez Oct 26 '11 at 21:21
I stand behind the term "well-defined" -- this is common usage. $f$ and the relation defining $f$ are commonly identified. – Bill Cook Oct 26 '11 at 21:21
@MarianoSuárez-Alvarez I don't follow your second comment. Do you mean that the implication fails to make sure that $f$ is defined for all elements of the domain? I wasn't claiming that "well-defined" is synonymous with "$f$ is a function". It's just part of the definition. – Bill Cook Oct 26 '11 at 21:24
I am saying that for the implication $x=y\implies f(x)=f(y)$ to be part of the definition of a relation being a function, then the symbol $f(x)$ has to be defined for all relations, not just functions (for otherwise the definition wouldbe circular) I claim that for any definition you give of what $f(x)$ means for a relation $f\subseteq X\times Y$ and an element $x\in X$, the implication will hold independently of whether the relation is or not a function. – Mariano Suárez-Alvarez Oct 26 '11 at 21:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.