If $\overline{L}$ is NL-hard, can we infer that $L$ is also NL-hard?

Question

NL denotes the class of languages, which can be decided by a non-deterministic Turing machine using $O(\log n)$ work tape cells.

A language $L$ is said to be NL-hard if $\forall L' \in$ NL : $L'\leq_L L\,$ (NL-hardness)

where $\leq_L$ denotes the log-space reduction, which is also transitive.

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Given a language L. Is the following claim true?

Claim: If $\overline{L}$ is NL-hard, then $L$ also NL-hard.

Given that $\overline{L}$ is NL-hard, we can log-space-reduce every problem contained in NL to $\overline{L}$, i.e. $\forall L'\in$ NL : $L'\leq_L \overline{L}$. Due to the Immerman–Szelepcsényi theorem we know that coNL = NL, but I do not know whether I can make progress with that fact.

Answer 1

Take some language $L'\in \text{NL}$, so since $\text{NL=coNL}$ we have $\overline{L'}\in \text{NL}$, and since $\overline{L}$ is $\text{NL}$-hard we have $\overline{L'}\leq_{log}\overline{L}$ therefore $L'\leq_{log}L$ (with the same reduction)