Proof of composition of one way function is not one way in general In this How to show composition of one way function is not such? question, the accepted answer uses g(x,y) as a one way function.
But, for a given output of g(x,y) = 0|x| of length l (say), isn't the output of g(any x of length l, 0l) the same as the earlier output? Isn't this a possible preimage of 0|x|?
"for a given output of g(x,y) = 0|x| of length l (say)," "the output of g(any x of length l, 0l)"

is "the same as the earlier output", which means that is "a possible preimage of 0|x|".

Accordingly, for ​ $g\hspace{-0.04 in}\circ \hspace{-0.03 in}f$ ​ to be one-way, it had better be the case that

$\operatorname{Prob}_{\langle \hspace{.02 in}x,y\rangle \hspace{.02 in}\leftarrow \hspace{.02 in}\left(\hspace{-0.03 in}\{\hspace{-0.02 in}0,1\hspace{-0.02 in}\}^{\hspace{.02 in}L}\hspace{-0.03 in}\right)^{\hspace{.05 in}2}}\hspace{-0.1 in}\left[g(x,\hspace{-0.04 in}y) = 0^{\hspace{.03 in}L}\hspace{-0.0 in}\right] \;\;\;$ is negligible.