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 ?

The direction $F(X_1)\
ot=F(X_2)\implies X_1\
ot=X_2$ 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 $$F(X_1)\
ot=F(X_2)\implies X_1\
ot=X_2;$$ the injective functions are (by definition) the ones which _also_ satisfy $$X_1\
ot=X_2\implies F(X_1)\
ot=F(X_2).$$