Let $f$ be a function whose domain is a set $X$. The function $f$ is said to be injective provided that for all $a$ and $b$ in $X$ if $f ( a ) = f ( b )$ , then $a=b$; that is, $f ( a ) = f ( b )$ implies $a = b$. Equivalently, if $a \ne b$ then $f ( a ) \ne f ( b )$ in the contrapositive statement.
Symbolically, $$\forall a , b \in X , f ( a ) = f ( b ) \implies a = b ,$$
which is logically equivalent to the contrapositive, $$\forall a , b \in X , a \ne b \implies f ( a ) ≠ f ( b ) .$$
Question. If I prove that $$f(a)\ne f(b)\implies a\ne b$$ can I conclude that $f$ is injective?
It doesn't work; a counterexample is given by $f(x) = \tan(x)$. Even if $f(a) \neq f(b) \implies a \neq b$, it is not an injective function, since it is $\pi$-periodic and thus an infinite number of points have the same image. You can convince yourself by looking at its graph.