injectivity question X1 ≠ X2 ⇒ F(X1) ≠ F (X2) I learned that in injectivity the following applies.
X1 ≠ X2 ⇒ F(X1) ≠ F (X2)
I was wondering if the other direction applies 
X1 ≠ X2 ⇐ F(X1) ≠ F (X2)
or is it  not possible to do the other direction since one needs to give the variables X1,X2 first? 
what if i said let X1,X2 be variables then follows for an injective function F that
 X1 ≠ X2  ⇐ F(X1) ≠ F (X2)
if thats true why did not my professor write a double headed arrow instead of one headed ? 
 A: The direction $F(X_1)\not=F(X_2)\implies X_1\not=X_2$ is a direct consequence of the definition of "function." A function can only take on one value per input, by definition. So $X_1=X_2\implies F(X_1)=F(X_2)$ for free, and this is the contrapositive of (hence, equivalent to) the property in question. That is:

Every function $F$ satisfies $$F(X_1)\not=F(X_2)\implies X_1\not=X_2;$$ the injective functions are (by definition) the ones which also satisfy $$X_1\not=X_2\implies F(X_1)\not=F(X_2).$$

A: Well, the implication $P\Rightarrow Q$ is logical equivalent to the contraposition $\neg Q\Rightarrow\neg P$, where $\neg$ means negation.
Here $x\ne y\Rightarrow f(x)\ne f(y)$ is logically equivalent to $f(x)=f(y) \Rightarrow x=y$.
The implication $f(x)\ne f(y)\Rightarrow x\ne y$ is part of the definition that $f$ is a function. Its called right uniqueness.
A function $f:A\rightarrow B$ is a relation $f\subseteq A\times B$ such that
(1) $f$ is left total: for each $a\in A$ there exists $b\in B$ with $(a,b)\in f$.
(2) $f$ is right unique: for each $a\in A$, $b,c\in B$, if $(a,b),(a,c)\in f$, then $b=c$.
