I read a commonly known proof of transitivity of logspace reduction: Lemma 6.: http://www.cs.au.dk/~arnsfelt/CT10/scribenotes/lecture5.pdf . In this proof it is used following trick: Let some turing machine $M_f$ for language $f$ compute only $j$-th bit of output. Ok, $M_f$ is logspace machine, however tape with output can require $O(n)$ space. Hence, we can't get $j$-th bit in such easy way.
Where am I wrong ?