Consider this snippet shown below from, An Introduction to Formal Languages and Automata 6th Edition by Peter Linz.
As per the text, choosing a value of $y = a^k$, where $k$ is odd is not permitted since this violates the condition of pumping lemma. This text simply says that the assumption on $k$ (number of $a$'s in $y$ being odd) is not permitted.
Now, as far as my understanding goes, the condition on choosing $xy$ is $|xy| \le p$, where $p$ is the pumping length, and $|y|>1$. And if we can show that one valid choice of $y$ satisfies the pumping lemma, we have been able to show that this language does not violate the pumping lemma, though the language can not be claimed to be regular. When I need to show that the language is not regular, I need to explore all possible $y$'s.
I fail to understand, why a $y$, with odd number of $a$'s, and $|y| \le p$, $x=\epsilon$, will not be permitted here.