January 26, 2025
These conditions are not very categoryish. How does one say m(a,m(b,c)) = m(m(a,b),c) without mentioning a, b, or c? Categories are just arrows and objects. Define i\colon M\to I\times M by i(a) = (\star,a).
We can define a functor on \mathbf{Set} by F(M) = I \times M, for any set M, where I is initial object of \mathbf{Set}, i.e., any singleton. Let α\colon F(M)\to M be the projection on the second component (\star,a)\mapsto a, a\in M.
If a\colon M\to N is and arrow the F(a)\colon F(M)\to F(N) by $IaM
An F-algebra on a category \mathbf{C} is a functor F\colon \mathbf{C}\to\mathbf{C}, an object A in \mathbf{C}, and an arrow α\colon F(A)\to A. For a monoid the F-algebra has functor F(M) = M\times M on \mathbf{Set} and arrow α\colon M\times M\to M where and (a,b)\mapsto m(a,b). Here I is an initial object of \mathbf{Set}, i.e., any singleton.
Sets can be equipped with an algebraic structure. For example, a semigroup is a set S with an associative binary operation. A binary operation m\colon S\times S\to S is associative if m(a,m(b,c)) = m(m(a,b),c), a,b,c\in M. If we write ab for m(a,b) this becomes a(bc) = (ab)c.
Exercise. The binary operation of subtraction on the integers is not associative.
A monoid is a semigroup with an identity element e satisfying ea = a = ae for all a in the semigroup.
Given an endofuction F\colon\mathbf{C}\to\mathbf{C}, an F-algebra \lbracket A,α\rbracket for category \mathbf{C} is an endofunctor F\colon\mathbf{C}\to\mathbf{C}, an object A, and an arrow α\colon F(A)\to A. If \lbracket B,β\rbraket is an F-algebra then f\colon A\to B is an F-algebra homomorphism if fα = βF(f).
If M is a moniod with identity e\in M then ea = a = ae, a\in M, and a(bc) = (ab)c, a,b,c\in M where ab = m(a,b) is the monoid operation m\colon M\times M\to M.
Need facts that are special to \mathbf{Set}.
B^A = \{f\colon A\to B\} = \{A\to B\}.
If f\colon X\to Y is a function, (x,x')\in\operatorname{ker}f iff f(x) = f(x') is an equivalence relation on X.
The map X\to X/\operatorname{ker}X, x\mapsto x/\operatorname{ker}f, is surjective
The map X/\operatorname{ker}f\to \operatorname{ran}f, x/\operatorname{ker}f\mapsto f(x), is well-defined and bijective
The map \operatorname{ran}f\to Y, f(x)\mapsto f(x), is injective.
f\colon X\to Y, f^\vdash\colon\mathcal{P}(Y)\to\mathcal{P}(X), f^\vdash(S) = \{x\in X\mid f(x)\in S\}\subseteq X, S\subseteq Y. x/\operatorname{ker}f = f^\vdash(\{f(x)\}).
Product \prod_{i\in I}A_i, \pi_i\colon \prod_{i\in I}A_i\to A_i such that p_i\colon C\to A_i implies there exists p\colon C\to\prod_{i\in I}A_i with \pi_ip = p_i all i\in I.
σ\colon J\to I, \Pi_{j\in J} A_{σ(j)}. Permutaion if σ is bijective. Projection if J\subseteq I. Reshape if J=n_1\times\cdots, (j_1,\ldots)\mapsto j_1 + n_1(\cdots).
τ\colon I\to J, \Pi_{j\in J} \Pi_{i\mid σ(i) = j} A_i. Grouping. If I = n and τ\colon n\to m, m \le n then I/\operatorname{ker}τ is a partition of I, τ_{0|12}\colon X_0\times X_1\times X_2\to (X_0)\times(X_\time X_2).
(A\times B)\times C \equiv A\times(B\times C)
\Pi_{i\in I} A \equiv A^I, π_i\colon \Pi_{i\in I} A\to A, π_i(a) = a(i). a\in\Pi_{i\in I} A\mapsto (i\mapsto π_i(a)).
Exponential e\colon (B^A\times A)\to B.
Product, A\to A\times B, and exponential, A\to B^A, are adjoint functors. \{(A \times B)\to C\} \equiv \{A \to \{B\to C\}\} f(a,b) = c iff (\underline{f}(a))(b) = c. (g(a))b = c iff \overline{g}(a, b) = c.
σ\colon I\to J, σ^\vdash J\to \operatorname{ker}σ.
M\times M\to M^2 product to exponential, as sets
M\times M\times M\to M\times M^2
M\times M\times M\to M^2\times M
\NN\to M\times\cdots\times M, σ\colon n\to m, m\le n.
M\times\cdots\times M\ to M^{σ(0)} \times M^b \times \ldots
m(a,m(b,c)) = m(m(a,b),c)
(a,b,c) -> (a, (b,c)) -> (a, m(b,c)) -> m(a, m(b,c))
m:G^2 -> G, m x m^2: G x G^2 ->
G^3-> G G^2 (a,b,c) -> (a,(b,c))
G^3-> G^2 G (a,b,c) -> ((a,b),c)
(a,b,c) -> (a,m(b,c))