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Possible Duplicate:
Conditions Equivalent to Injectivity

Let $A$ := "$f$ is injective" and $B$ := "$f(X \cap Y) = f(X) \cap f(Y)$".

My first idea is to show $B \implies A$ through contraposition, so $\lnot A \implies \lnot B$. Would it then be enough if I say: $f$ is not injective and then show an example where the equation in $B$ is wrong? Would it be a proof then?

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    Jay: If you mean that **for every non-injective $f$** you find an example of sets $X$, $Y$ for which the equality fails, then yes; this is a correct proof.2011-12-09

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