The unit inserts generators as one-letter words and the counit evaluates words in the free-group adjunction

Under [L1], the identity of $F(X)$ transposes to the generator inclusion, so ${\eta }_X(x)$ is the one-letter word $x$. The identity function of $U(G)$ extends uniquely to ${\epsilon }_G$, hence ${\epsilon }_G$ evaluates a word by multiplying its letters in $G$.

On a generator $x$, the composite ${\epsilon }_{F(X)}F({\eta }_X)$ first makes the one-letter word whose letter is the word $x$, then evaluates it to $x$. The two homomorphisms agree on all generators, so the first identity in [F2] holds.

For $g\in G$, the composite $U({\epsilon }_G){\eta }_{U(G)}$ forms the one-letter word $g$ and evaluates it to $g$. Thus the second identity in [F2] holds.

If $X=\mathrm{\varnothing }$, the first check is equality of the unique homomorphisms from the trivial free group. If $G$ is trivial, every evaluated word is its identity. Thus both boundary cases obey the same formulas.