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?

  • $\begingroup$ What is $S$ here? the argument of the functor? $\endgroup$
    – Berci
    Oct 3 '17 at 22:41
  • $\begingroup$ @Berci, $S^2$ is a second power functor which maps $X$ to $X\times X$. $\endgroup$
    – A. G
    Oct 3 '17 at 22:43
  • $\begingroup$ Also, is $T$ a monad, or simply an endofunctor? $\endgroup$
    – Berci
    Oct 3 '17 at 22:55
  • $\begingroup$ @Berci It is an endofunctor. $\endgroup$
    – A. G
    Oct 3 '17 at 22:58

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.

  • $\begingroup$ so, in the second case we map $1$ to an arbitrary element of $X$? $\endgroup$
    – A. G
    Oct 3 '17 at 23:00
  • $\begingroup$ Yes. How does the definition of morphism of $T$-algebras translate here? $\endgroup$
    – Berci
    Oct 3 '17 at 23:11
  • $\begingroup$ Sorry, I didn’t get your thought. What do you mean by your question? @Berci $\endgroup$
    – A. G
    Oct 3 '17 at 23:18

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