0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.