A map is one-to-one or injective if $a_1 \neq a_2 $ implies $f(a_1) \neq f(a_2)$. Equivalently, a function is one-to-one if $f(a_1) = f(a_2)$ implies $a_1 = a_2$.

Can I define it in the below mentioned way?

A map is one-to-one or injective if $f(a_1) \neq f(a_2)$ implies $a_1 \neq a_2 $. Equivalently, a function is one-to-one if $a_1 = a_2$ implies$f(a_1) = f(a_2)$ .

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    $\begingroup$ NO; an Injective function "is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain." Your proposed def, is simply the condition of "functionality": the same element has not two difefrent images. $\endgroup$ – Mauro ALLEGRANZA Nov 12 '17 at 13:40

You cannot use this. Any function satisfies this, because if $a_1 = a_2$, then $f(a_1)$ is the same as $f(a_2)$, because what this notation "means" intuitively, that you evaluate the function $f$ in the point $a_1$. But since $a_1 = a_2$, if you evaluate your function in $a_2$ you are actually evaluating the function in $a_1$, so they will result in the same result.

If you know about relations, what you have is that $a_1 R f(a_1)$, since $a_1 = a_2$ you also have $a_2 R f(a_1)$, and by definition we also have $a_2 R f(a_2)$. But because functions are the relations so that for all points $x$ there is one unique point $f(x)$ such that $x R f(x)$. This means that $f(a_1) = f(a_2)$


No, you definitely can't change the order! If you have two statements $A$ and $B$ then $$ (A\Rightarrow B) \Leftrightarrow (\neg B\Rightarrow \neg A). $$ But $A\Rightarrow B$ is not equivalent to $B\Rightarrow A$. You can never do it for any statements $A$ and $B$! Never!

In your case you can see it on the not injective function $f(x)=x^2$. It isn't injective since $f(1)=1=f(-1)$ and $1\neq -1$. But for $f(x)\neq f(y)$ you can conclude $x\neq y$.


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