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Suppose that $f:X \rightarrow Y$ is a function.

Then an injection can be defined as:

$\forall x_1,x_2 \in X, f(x_1) = f(x_2) \Rightarrow x_1=x_2$

Why isn't it defined instead as follows:

$\forall x_1,x_2 \in X, f(x_1) = f(x_2) \Leftrightarrow x_1=x_2$

I think the above statement also captures the situation described by the definition of an injection in words, which is:

If no element of $Y$ is assigned to more than one elementof $X$, i.e. the function takes a different value for each point of the domain.

I can see that the definition is missing the phrase "if and only if". So is that the only reason we don't write "$\Leftrightarrow$"?

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It could be defined like that, the but the other direction is always true by definition of function. –  Git Gud Jul 11 at 13:23
    
Your question has been answered adequately. Here an alternative definition: $f$ is injective if it is left-cancellable; i.e. $f\circ g=f\circ h$ implies that $g=h$. –  drhab Jul 11 at 13:59

3 Answers 3

up vote 3 down vote accepted

Recall that the fact that $f$ is a function implies that $x_1=x_2\implies f(x_1)=f(x_2)$.

So if you say that $f$ is an injective function, then $\leftarrow$ part is already true. So in order to save ourselves trivialities, we only state the $\implies$ part, and then when you want to prove that $f$ is injective you don't have to worry about that.

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Here is another definition. $$(\forall x_1,x_2 \in X, f(x_1) = f(x_2) \Leftrightarrow x_1=x_2) \wedge (1+1 = 2)$$ –  user147887 Jul 11 at 13:27
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What's your point? –  Asaf Karagila Jul 11 at 13:39
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@AsafKaragila The point, I believe, is to show the OP that one can include a bunch of needless trivialities in the definitions. –  Git Gud Jul 11 at 13:41
    
@GitGud: I see. Yes, that is a point. But it seems to interprets the question as being facetious, rather than an actual question of a newbie. –  Asaf Karagila Jul 11 at 13:46
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I didn't mean to seem facetious, I just wanted to emphasize that the reason it is not "if and only if" is not because it is wrong but because it is unnecessary. –  user147887 Jul 11 at 13:51

Both of the definition you propose are equivalent if $f$ is function. By definition if $f$ is a function, $x_1=x_2$ implies $f(x_1)=f(x_2)$.

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@GitGud Yeah sorry, it was not very well expressed. –  Surb Jul 11 at 13:36
    
No need to apologize. I removed my comment. –  Git Gud Jul 11 at 13:37

If you suppose $f$ as a function, than the converse of your implication is not necessary. The implication $x_1=x_2 \implies f(x_1)=f(x_2)$ tells us only that f is a function!

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"The implication $x_1=x_2 \implies f(x_1)=f(x_2)$ tells us only that $f$ is a function!" This implication already assumes $f$ is a function, other wise it doesn't make sense. –  Git Gud Jul 11 at 13:33
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I mean that $x_1=x_2 \implies f(x_1)=f(x_2)$ is something that only makes sense if $f$ is already known to be a function. Well, if $f$ is already known to be a function, you don't need $x_1=x_2 \implies f(x_1)=f(x_2)$ to claim that $f$ is a function. And if it is not known whether $f$ is a function or not, then $x_1=x_2 \implies f(x_1)=f(x_2)$ is a meaningless string of symbols. –  Git Gud Jul 11 at 13:40
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Why not? f(x) may be a set of elements! –  Emin Jul 11 at 13:52
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@drhab, actually Emin is correct. The notation $f(x)$ for a relation $f$ is very commonly used to denote the set of all elements related to $x$, or even in many circumstances just one possible element related to $x$. For instance, this is common when working with canonical relations in symplectic geometry or mathematical physics. –  Santiago Canez Jul 11 at 14:15
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@SantiagoCanez The standard notation for that would be $f[\{x\}]$. I reiterate, when $f$ happens to be a function, if you interpret $f(x)$ as you suggest, you'll be getting $f(x)=\{a\}$ and $f(x)=a$, for some $a$. And to be perfectly honest, the fact that geometers and physicists use that notation, only supports my claim that it is incorrect since they tend to abuse notations. –  Git Gud Jul 11 at 14:27

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