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I'm new to category theory, and I'm trying to define two categories which are not isomorphic, yet have identical graphs.

  1. Category $A$ has 3 objects: {0, 1, 2}. For each pair of objects $m, n \in A$ there is a morphism $f_{m} : m \xrightarrow{f_{m}} n$ iff $(m + m) \% 3 = n$. Each object additionally has a morphism $id_{m} : m \xrightarrow{id_{m}} m$. Composition $f_{n} \circ f_{m} = id_{m}$ whenever the codomain of $f_{m}$ is equal to the domain of $f_{n}$.

  2. The objects of category $B$ are 3 kinds of flowers: {"Bluebell", "Texas Bluebonnet", "Red Poppy"}. For each pair of (not necessarily distinct) flowers $p, r \in B$ there is a morphism $h_{pr} : p \xrightarrow{h_{pr}} r$ iff $p$ and $r$ have the same color petals (as implied by their names). The composition $h_{qr} \circ h_{pq}$ is equal to $h_{pr}$.

  3. Functor $F : A \rightarrow B$ maps objects $F(2) = $"Bluebell", $F(1) =$ "Texas Bluebonnet", and $F(0) = $"Red Poppy". It also maps identities of $A$ according to $F(id_{m}) = h_{F(m)F(m)}$, where $m \in A$, and maps the modulo morphisms of $A$ according to $F(f_{m}) = h_{F(m)F(n)}$ where $n = (m + m) \% 3$.

  4. Functor $G : B \rightarrow A$ maps objects $G($"Bluebell"$) = 2$, $G($"Texas Bluebonnet"$) = 1$, and $G($"Red Poppy"$) = 0$. It also maps morphisms of B according to $G(h_{pp}) = id_{G(p)}$, and $G(h_{pr}) = f_{G(p)}$, where $p,r \in B$ and $p \neq r$.

The categories are not isomorphic, because $F(f_{0}) = F(id_{0}) = h_{BB}$, while $G(h_{BB}) = id_{0}$ (abbreviating "Bluebell" as $_{B}$). But the graphs should still be the same, because identity morphisms do not appear in the graph of a category. Is this correct?

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The definitions of $A$ and $B$ are unclear. Please elaborate. – Martin Brandenburg Apr 18 '12 at 8:32
Can you tell me what is unclear? I revised the whole question several times before posting, so it includes all the details I can think of. Please let me know what is missing. – Byron Hawkins Apr 18 '12 at 11:29
@Byron: I'm writing an answer that will hopefully clear things up. – joriki Apr 18 '12 at 11:29
up vote 4 down vote accepted

The definitions of the categories are fundamentally flawed. I suggest that you go back to the definition of a category and try to understand why what you've written down doesn't work out. A morphism is an arrow from one object to another, so "its only morphism is addition modulo $3$" doesn't make sense: addition modulo $3$ assigns a third number to two numbers, which isn't what we need to define a morphism. The same for path composition, which assigns a third path to two paths.

Here are three categories that have some similarity with what you've written; perhaps one or two of them are what you were trying to get at.

Category $A'$ has one object, the set $\mathbb Z_3=\{0,1,2\}$. For every $n\in\mathbb Z_3$, there is one morphism from $\mathbb Z_3$ to $\mathbb Z_3$, namely the function that sends $x$ to $x+n\bmod3$. Composing the morphisms corresponding to $n_1$ and $n_2$ yields the morphism corresponding to $n_1+n_2\bmod3$. The identity morphism for the only object is the one corresponding to $n=0$.

Category $A''$ has three objects, $0$, $1$ and $2$. There is one morphism for each pair of objects. Composition yields the only morphism between the appropriate objects, and the identity morphism for an object is the only morphism from that object to itself.

Category $B'$ has three objects, $a$, $b$ and $c$. There is one morphism from $x$ to $y$ for each path from $x$ to $y$ in the complete graph on $\{a,b,c\}$. Composition of morphisms is defined by composition of the corresponding paths. The identity morphism for an object is the path beginning and ending at that object without any edges.

Now we can define a functor $F$ from $B'$ to $A''$ that assigns to the objects $a$, $b$, $c$ the objects $0$, $1$, $2$, respectively, and to the morphism corresponding to a path from $x$ to $y$ the only morphism from $F(x)$ to $F(y)$. We can also define a functor $G$ from $A''$ to $A'$ that assigns to each object in $A''$ the only object in $A'$, and to the morphism from $x$ to $y$ the morphism that maps $x$ to $y$. We can also define the composition $H=G\circ F$ from $B'$ to $A'$.

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Yes, $A''$ and $B'$ are precisely what I was meaning to define. Given these corrections, can we say that $F$ and $F^{-1}$ are isomorphic functors? It's unclear to me whether $F \circ F^{-1} = id_{A''}$. In particular, I don't understand exactly how one ought to define $F(id_{A''})$. Perhaps it depends on how $id_{A''}$ is defined, and there seem to be 2 options: 1. $id_{A''}(x) = x$ 2. $id_{A''}(x) = (x + 0) \% 3$ (i.e., it is selected from among the other $A''$ morphisms) – Byron Hawkins Apr 18 '12 at 16:07
By the way, I don't actually see the difference between your definitions of $A''$ and $B'$ and my definitions of $A$ and $B$. I didn't intend to introduce any third object in the morphisms. Can you help me understand what is incorrect about the way I defined them? – Byron Hawkins Apr 18 '12 at 21:51
@Byron: A lot of your first comment is again fundamentally flawed. a) No $F^{-1}$ has been defined, and I don't see how to define one. b) Your use of $\operatorname{id}_{A''}$ is inconsistent. In $F \circ F^{-1} = \operatorname{id}_{A''}$ it seems to be a functor whereas in $F(\operatorname{id}_{A''})$ it seems to be a morphism. c) I don't see the difference between 1. and 2., since $(x+0)\%3=x$. About your second comment: But addition modulo $3$ does introduce a third object. I think you should spell out your definition of $A$; that would make it easier to see what's wrong with it. – joriki Apr 18 '12 at 23:21
@Byron: You're absolutely right to start out with these simple examples. I wouldn't say that in general the objects "ought to be" structures instead of atomic units -- it's perfectly usual to consider objects without internal structure e.g. in a poset category‌​. Regarding your new definitions, I don't understand them. First, for $A$, you define six morphisms, one for each ordered pair of distinct objects. This would preclude the existence of identity morphisms, since those have to be from an object to itself. – joriki Apr 19 '12 at 2:01
@Byron: You're welcome -- I'm glad to hear it paid off in the end :-) – joriki Apr 20 '12 at 15:48

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