Most proofs of the problem that I have seen stop at proving $\textbf{NL}\subseteq \textbf{NC}^2$ and not explicitly point out the circuit. I'm trying to show that $\textbf{NC}$ circuit family can compute log space reduction by constructing corresponding $\textbf{NC}^2$, using the definition of implicitly logspace computable function according to Arora and Barak:
A function $f : \{0, 1\}^∗\rightarrow \{0, 1\}^∗$ is implicitly logspace computable, if $f$ is polynomially bounded (i.e., there’s some $c$ such that $|f(x)|\le |x|^c$ for every $x\in \{0, 1\}^∗$ ) and the languages $L_f = \{\langle x, i\rangle | f (x)_i = 1\}$ and $L'_f = \{\langle x, i\rangle | i \le |f(x)|\}$ are in $\textbf{L}$.
Where $f(x)_i$ is the $i$th bit of $f(x)$. In my opinion, we can compute $i$th bit of $f(x)$ and determine the length of $f(x)$ by logspace machines.
Because $L_f\in\textbf{L}$, there is a $\textbf{NC}^2$ circuit family decides $L_f$. $f(x)_i$ is computed by a circuit $C_{|\langle x,i\rangle|}(\langle x, i\rangle)$. We could generate all $f(x)$ bits by merging $|f(x)|=|x|^c$ circuits $C_{|\langle x,i\rangle|}$ for $i=0,\ldots,|x|^c$ "hard-wired" their input gates for $i$ in binary encode, such as $C(\langle x,0\rangle),C(\langle x,1\rangle),C(\langle x,10\rangle),$ etc. The final circuit has $O(\log^2n)$ depth and $poly(n)$ size because each circuit component also has $O(\log^2n)$ depth and $poly(n)$ size.
Does my proof have any defect?