Dtyjlui

Let Set denote the category of sets.

Let $T:$ Set $to$ Set be the functor that sends a set $X$ to the set of finite words on $X$.

That is, $TX = {[x_m,..,x_1] : m = 0,1,2,3..., x_i in X}$

$T$ can be considered a monad on Set with multiplication given by concatenation, and the unit by $x to [x]$.

I'm trying to show explicitly that the category of $T$ algebras on Set
is equivalent to the category of monoids (unital).

To do this we can define a functor $F:$ Mon $to$ Alg$_T$(Set) by $F(m) = (m,a)$ where $a: Tm to m$ is s.t $a([f_n,..,f_1]) = f_n...f_1$.

I have shown that this functor is fully faithful.

However I'm not sure I understand if it is essentially surjective.

Letting $(X,a_X) in$ Alg$_T$(Set). I thought of defining a monoid $m$ as $TX$, where the action is concatenation, and the unit is simply the empty word.

Doing this yields a natural map in Alg$_T$(Set) from $TX$ to $X$, namely $a_X$. Clearly this map is onto because $a_X([x]) = x$ by definition of $(X, a_X)$ being an algebra.

But is this map injective?

You're starting from a $T$-algebra $(X,a_X)$, and you wish to show that it is isomorphic to $F(M)$ for some monoid $M$.

In particular, since $F(M)$ is of the form $(M,h)$ for $h:TMto M, [m_1,...,m_n]mapsto m_1...m_n$, then if there is such an isomorphism $f:Xto M$ and if $x,yin X$, then $hcirc T(f)([x,y]) = h([f(x),f(y)])=f(x)f(y)$ and $hcirc T(f)([x,y] ) = fcirc a_X([x,y])$.

Considering that $f$ is a bijection and thus identifying $M$ and $X$ with it, we get a monoid structure on $X$ such that $a_X([x,y]) = xy$.

Thus now if we want to find $M$, we know what we should do : define multiplication on $X$ by the formula $xy := a_X([x,y])$; and check that this makes $X$ into a monoid such that $F(X) simeq (X,a_X)$ (actually it will be an equality !)

To prove that $X$ is indeed a monoid with this operation, you will need to use the different properties of the monad and axioms of $T$-algebra.

Associativity will follow from the associativity axiom of $T$-algebras that says that this should commute :

$require{AMScd} begin{CD}
TTX @>{mu_X}>> TX\ @V{T(a_X)}VV @VV{a_X}V\
TX @>>{a_X}> X
end{CD}$

and the unit will be $a_X([ ])$ and the fact that this is a unit will follow from the associativity axiom for $T$-algebras, together with the unit axiom of $T$-algebras that says that this should commute :

$require{AMScd} begin{CD}
X @>{eta_X}>> TX\ @V{id_X}VV @VV{a_X}V\
X @>>{id_X}> X
end{CD}$

You then need to prove that $a_X([x_1,...,x_n]) = x_1...x_n$ but this will follow from associativity again. I'll let you check the details.