Let $L = \emptyset$ and $L' = \{a\}$ be two languages over an arbitrary non-empty alphabet $\Sigma$, $a \in \Sigma$.

$L$ can be reduced to $L'$: the reduction just transforms anything it is given into the word $aa$.

There is no way to reduce $L'$ to $L$: the answer would be always negative.

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    $\begingroup$ Yes, your proof is fine. $\endgroup$ – David Richerby Mar 16 '19 at 18:32
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    $\begingroup$ We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. $\endgroup$ – D.W. Mar 16 '19 at 18:45
  • $\begingroup$ No, the proof is not fine as a problem is understood as not empty. $\endgroup$ – John L. Mar 16 '19 at 18:52
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    $\begingroup$ @Apass.Jack I don't understand what you're trying to say. $\emptyset$ reduces to any language except $\Sigma^*$, since you just map every instance to some fixed, chosen "no" instance of the target. But no non-empty language reduces to $\emptyset$ because there's nothing to map the "yes" instances to. And that's what the asker says. $\endgroup$ – David Richerby Mar 16 '19 at 20:47

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