The canonical algebra presentation is split in the base, but its canonical splittings need not be algebra homomorphisms

The algebra law gives $aT(a)=a{\mu }_A$ and $a{\eta }_A=1_A$. The monad unit law gives ${\mu }_A{\eta }_{TA}=1_{TA}$, while naturality of $\eta$ gives $T(a){\eta }_{TA}={\eta }_{A}a$. These are exactly the four equations of [L2] for $t={\eta }_{TA}$ and $s={\eta }_A$.

For the free-monoid monad, take the monoid $M=\{1,e\}$ with $e^2=e$. On the two-letter word $[e,e]$, the composite ${\eta }_{M}a$ first multiplies and gives the one-letter word $[e]$, whereas ${\mu }_{M}T({\eta }_M)$ gives the two-letter word $[e,e]$.

Therefore the underlying canonical presentation is split in $\mathcal{C}$.

The equality ${\eta }_{M}a={\mu }_{M}T({\eta }_M)$ is precisely the algebra-homomorphism equation for ${\eta }_M:(M,a)\to (TM,{\mu }_M)$, and step 1.2 shows it fails.

For that same algebra no algebra section exists at all. Under the isomorphism over $\mathbf{S}\mathbf{e}\mathbf{t}$ of [L3], an algebra map $s:(M,a)\to (TM,{\mu }_M)$ with $as=1_M$ is a monoid homomorphism $\sigma$ from $M$ to the free monoid on the set $M$ whose composite with word evaluation is the identity. Concatenation adds word lengths, so $w^2=w$ forces $\mathrm{\mid }w\mathrm{\mid }=0$ and the empty word is the only idempotent of that free monoid; since $e^2=e$ in $M$, $\sigma (e)$ is the empty word and evaluates to $1\mathrm{\ne }e$. Hence the presentation of $(M,a)$ is not split in ${\mathcal{C}}^T$.

Thus the canonical splittings always exist in the base by step 2.1, the canonical ones need not lift to algebra homomorphisms by step 2.2, and by step 2.3 the presentation itself need not be split in ${\mathcal{C}}^T$. No failure is asserted for every monad or every algebra.