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When we prove that a mapping $f: X \rightarrow Y$ is one-to-one (injective), we normally show that $\forall a, b \in X$, if $f(a) = f(b)$ holds, then $a=b$. Alternatively, by contrapositive, we could show $\forall a, b \in X$, $a \neq b$ implies $f(a) \neq f(b)$.

May I ask that is it correct to show that $\forall a, b \in X$, if $a=b$ holds, then $f(a) = f(b)$? If it is wrong, could you give me a counterexample? Thanks

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  • $\begingroup$ $a=b\implies f(a)=f(b)$ is one of the definitiory properties of "being a function". In fact, it is kind of the thing that justifies writing "$f(a)$" in the first place. So, no, you are indeed wrong. $\endgroup$ – user228113 Mar 1 '17 at 23:15
  • $\begingroup$ counter example: just take any constant function $f:\mathbb{N} \to \mathbb{N}$, it is not injective but your argument works $\endgroup$ – Saravanan Mar 1 '17 at 23:22
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The given condition "$ a = b \implies f(a) = f(b) $ holds for all $ a, b \in X $" is technically correct, but it is not used meaningfully to show injectivity.

For example, consider the sets $ X = \{-1,1\}, Y = \{1\} $ and the function $ f: X \to Y $ that maps $ X $ to $ Y $ with $ f(x) = x^2 $. Then for any $ x \in X $, we have $ f(x) = 1 $ and clearly the condition $ a = b \implies f(a) = f(b) $ holds for all $ a, b \in X $. And yet this map is clearly not injective.

Of course while the given condition is not particularly useful regarding injectivity, showing uniqueness is another matter altogether.

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  • $\begingroup$ Is there an example that a correspondence $f : X \rightarrow Y$ is injective, but this correspondence does not satisfy the condition that " $a = b$ implies $f(a) = f(b)$"? $\endgroup$ – Paradiesvogel Mar 2 '17 at 0:59

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