If $X \subseteq B$ and $f: A \rightarrow B$ then prove the above. Consider that $X$, $A$ and $B$ are all sets.
If $f$ is injective this would be solved easily but the problem is that $f$ is not necessarily injective.
If $X \subseteq B$ and $f: A \rightarrow B$ then prove the above. Consider that $X$, $A$ and $B$ are all sets.
If $f$ is injective this would be solved easily but the problem is that $f$ is not necessarily injective.
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