# Categories of $\mathbf T$-algebras.

What is the way of describing Categories of the form $\mathbf A\mathbf l\mathbf g(T)$ for functors, for example:

constant functor $T:\mathbf S\mathbf e\mathbf t\rightarrow \mathbf S\mathbf e\mathbf t$ with
1) value $\emptyset$;
2) value $1=\{0 \}$,
3)$$S^2:\mathbf S\mathbf e\mathbf t\rightarrow \mathbf S\mathbf e\mathbf t. How do I figure out what the algebras for a given functor look like? • What is S here? the argument of the functor? Oct 3 '17 at 22:41 • @Berci, S^2 is a second power functor which maps X to X\times X. – A. G Oct 3 '17 at 22:43 • Also, is T a monad, or simply an endofunctor? Oct 3 '17 at 22:55 • @Berci It is an endofunctor. – A. G Oct 3 '17 at 22:58 ## 2 Answers Let us consider polynomial functors, which are functors that we can build up from the following operations: • constant functors • cartesian product \times • disjont sum + For example, such a functor might be$$T(X) = C \times X + X \times X \times (D + X \times X)$$where C and D are fixed sets. Take your functor and use distributivity to write it out as a "polynomial", i.e., a disjoint sum of products. Let us also write X^n for the product X \times X \times \cdots \times X of n copies of X. The above functor T would be$$T(X) = C \times X + D \times X^2 + X^4.$$An algebra for T is a set A together with a structure map$$a : T(A) \to A$$Because we expressed T(A) as a sum of powers, such an a is equivalent to having several maps. For example,$$a : C \times A + D \times A^2 + A^4 \to A$$is equivalent to three maps$$\begin{align*} a_1 &: C \times A \to A \\ a_2 &: D \times A^2 \to A \\ a_3 &: A^4 \to A \end{align*}$But these are precisely the operations for our algebra! This procedure works always. Here are some examples. Take$T(X) = \emptyset$. Then a$T$-algebra is a set$A$with a map$a : \emptyset \to A$. But there is exactly one such map which is not doing anything, so the answer is that a$T$-algebra is just a set. Take$T(X) = 1$. Then a$T$-algebra is a set$A$with a map$a : 1 \to A$. Such a map is the same thing as one element of$A$, so the answer is that a$T$-algebra is a set$A$together with one element. This is known as pointed set. Take$T(X) = X \times X$. A$T$-algebra is a set$A$with a map$a : A \times A \to A$, i.e., a set with a binary operations. This is known as magma. Just for fun, let us try one more. Take$T(X) = 1 + X + X^2$. A$T$algebra is a map$a : 1 + A + A^2 \to A$which is equivalent to having • a map$a_1 : 1 \to A$, which is the same as having one element of$A$, and • a map$a_2 : A \to A$, which is just a unary operation on$A$, • a map$a_3 : A^2 \to A$, which is a binary operation. Thus, a$T$-algebra is a set$A$together with one element, one unary operation, and one binary operation. The way is to unfold the definition. A$T$-algebra is a pair$(X,f)$where$f$is the 'operation'$T(X)\to X$. Now, for (1), the operation is the unique (empty) function$\emptyset\to X$, so it basically adds no more structure, and we can get an isomorphism${\bf Alg}(T)\cong{\bf Set}$by adding/removing the empty function [$A\mapsto (A,\underset{\emptyset\to A}0)\mapsto A$]. For (2), we will get the category of pointed sets, i.e. sets with a distinguished point, and morphisms must respect the distinguished points. This is because the possible 'operations'$1\to X$now are just picking an element of$X$. For (3), we get the category of magmas, i.e. a set$X$equipped with any binary algebraic operation - that is, an arbitrary function$X\times X\to X$. Morphisms between magmas must respect their given operations. Note for example that the category of semigroups - where we also require the binary operation to be associative - is a full subcategory of that of magmas. • so, in the second case we map$1$to an arbitrary element of$X$? – A. G Oct 3 '17 at 23:00 • Yes. How does the definition of morphism of$T\$-algebras translate here? Oct 3 '17 at 23:11
• Sorry, I didn’t get your thought. What do you mean by your question? @Berci
– A. G
Oct 3 '17 at 23:18