 Solution

A logspace transducer is a deterministic Turing machine with a read-only input tape, $O(\log (n+2))$ read-write work space, and a write-only output tape. It emits symbols successively and cannot reread the output; that output is not charged as working memory. For a total function it must halt on every input. A logspace-computable function $f$ is a total function whose output $f(x)$ is produced by such a logspace transducer on input $x$.

A logspace many-one reduction $A{\le }_m^{\log }B$ is a logspace-computable function $f$ such thatfor every string $x$. The logarithmic space bound permits a polynomial-length output, even when that output does not fit in work space. Indeed, a halting logspace transducer cannot repeat its state, input-head position and work-tape contents and head positions: the unreadable output cannot affect its subsequent behaviour. There are only a polynomial number of such internal Turing machine configurations, hence a polynomial bound on the number of steps and output symbols.

To prove a formal language $B$ is NL-complete, first give an NL acceptance procedure for $B$. Then show every $A\in \mathrm{NL}$ has a logspace many-one reduction to $B$. Alternatively, reduce one already NL-complete formal language to $B$, using transitivity of logspace many-one reductions.

Transitivity requires logspace composition by recomputation, rather than storing the intermediate output. To compose $g$ with $f$, simulate the logspace transducer for $g$ with a virtual input $f(x)$. Retain its virtual input-head position in $O(\log n)$ bits. Whenever it asks for a symbol, rerun the logspace transducer for $f$ from the beginning and count its output symbols until that position is reached. Since $|f(x)|$ is polynomial in $n$, the simulated work space $O(\log |f(x)|)$ and all output-position counters are $O(\log n)$. Repeated runs reuse their storage. This proves closure under composition and explains why the hardness argument works.