**Puzzle 98**: As alluded to in my solution to Puzzle 97, a category with just one object is equivalent to a monoid.

**Puzzle 99**: A cute way to keep track of morphisms is to imagine placing the square on a table, drawing numbers on each corner of the square, and drawing matching numbers on the table next to the square. (Draw the same numbers on the same corners of the back face, as though the square were see-through or something. We're just labeling vertices.) Now every possible placement of the square on the table (including passing it through a mirror first) corresponds to a permutation map, where each number on the table maps to the adjacent corner on the square.

[As Keith mentioned](https://forum.azimuthproject.org/discussion/comment/18690/#Comment_18690), the symmetries of the square are given by the dihedral group \\(D_8\\) (using the algebraic indexing convention). The permutations I described above correspond to the elements of this group (remember, [every group is a permutation group](https://en.wikipedia.org/wiki/Cayley%27s_theorem)). Since categories with one object are just monoids, we expect that our construction here ought to be a monoid -- and indeed, groups are just monoids with inverses.

There are eight morphisms in this category, whence the name \\(D_8\\). This category is _not_ a preorder, since in a preorder, there is at most one morphism between any pair of objects. This would necessitate that the only endomorphism on an object must be the identity, which is not the case here.

**Puzzle 100**: We have all the same permutations as in the case of \\(D_8\\), meaning every object will have eight morphisms coming out of it. Because this is a group, every object also has eight morphisms coming into it -- just take inverses.

Consider the effect of any particular permutation in \\(D_8\\) on the colored triangle. An assignment of the colored triangle to a location also uniquely determines the rest of the square; hence, if any pair of permutations send the triangle to the same place, they must be the same. So there is a unique morphism between any two objects \\(X\\) and \\(Y\\), since every object has a unique placement of the triangle. Hence, this category is just the codiscrete preorder on eight objects. This also means that there are \\(8 \cdot 8 = 64\\) morphisms in this category.

**Puzzle 101**: The requirement that a set \\(S\\) have at most one function from any particular set to \\(S\\) necessitates that \\(S\\) be either a singleton set or the empty set. A larger set would support at least two constant functions from any other set. On the other hand, the requirement that there be at most one function out of \\(S\\) into any particular set mandates that \\(S\\) be the empty set, since any larger set would support multiple functions into any set of size at least two. So the only candidate is the empty set, which has no maps in and the trivial, "impossible" map out for every other set. (Which _does_ count as the identity on the empty set.)

Not only is this subcategory a preorder, it's also a partial order. And a total order.

**Puzzle 99**: A cute way to keep track of morphisms is to imagine placing the square on a table, drawing numbers on each corner of the square, and drawing matching numbers on the table next to the square. (Draw the same numbers on the same corners of the back face, as though the square were see-through or something. We're just labeling vertices.) Now every possible placement of the square on the table (including passing it through a mirror first) corresponds to a permutation map, where each number on the table maps to the adjacent corner on the square.

[As Keith mentioned](https://forum.azimuthproject.org/discussion/comment/18690/#Comment_18690), the symmetries of the square are given by the dihedral group \\(D_8\\) (using the algebraic indexing convention). The permutations I described above correspond to the elements of this group (remember, [every group is a permutation group](https://en.wikipedia.org/wiki/Cayley%27s_theorem)). Since categories with one object are just monoids, we expect that our construction here ought to be a monoid -- and indeed, groups are just monoids with inverses.

There are eight morphisms in this category, whence the name \\(D_8\\). This category is _not_ a preorder, since in a preorder, there is at most one morphism between any pair of objects. This would necessitate that the only endomorphism on an object must be the identity, which is not the case here.

**Puzzle 100**: We have all the same permutations as in the case of \\(D_8\\), meaning every object will have eight morphisms coming out of it. Because this is a group, every object also has eight morphisms coming into it -- just take inverses.

Consider the effect of any particular permutation in \\(D_8\\) on the colored triangle. An assignment of the colored triangle to a location also uniquely determines the rest of the square; hence, if any pair of permutations send the triangle to the same place, they must be the same. So there is a unique morphism between any two objects \\(X\\) and \\(Y\\), since every object has a unique placement of the triangle. Hence, this category is just the codiscrete preorder on eight objects. This also means that there are \\(8 \cdot 8 = 64\\) morphisms in this category.

**Puzzle 101**: The requirement that a set \\(S\\) have at most one function from any particular set to \\(S\\) necessitates that \\(S\\) be either a singleton set or the empty set. A larger set would support at least two constant functions from any other set. On the other hand, the requirement that there be at most one function out of \\(S\\) into any particular set mandates that \\(S\\) be the empty set, since any larger set would support multiple functions into any set of size at least two. So the only candidate is the empty set, which has no maps in and the trivial, "impossible" map out for every other set. (Which _does_ count as the identity on the empty set.)

Not only is this subcategory a preorder, it's also a partial order. And a total order.