A monic coalgebra morphism whose underlying $\text{Set}$ morphism is not injective

Question

Let $F:\mathscr A\to \mathscr A$ be a functor. Consider the following category $\mathscr C$. The objects are arrows $A\to F(A)$. If $\alpha:A\to F(A)$ and $\beta:B\to F(B)$ are two objects, then a morphism from $\alpha$ to $\beta$ is an arrow $h:A\to B$ such that $F(h)\circ \alpha=\beta\circ h$.

Now let $F: Set\to Set$ be the functor that assigns to every set $S$ the subset of $S^3$ consisting of all triples in which at least two elements are the same. Let $A=\{a,b,c\}$ and let $\alpha$ act as follows: $a\mapsto (a,a,b);b\mapsto (a,a,a); c\mapsto(a,b,b)$. The problem is to find an object $\beta:B\to F(B)$ of $\mathscr C$ and a morphism $ \alpha\to\beta$ in $\mathscr C$ such that the morphism is monic but the underlying set function $A\to B$ is not injective.

I tried to play with some examples. For example, let $B=\{1\}$. Then there is only one possible $\beta:B\to F(B), 1\mapsto (1,1,1)$. And there is only one possible morphism $\alpha\to \beta$ that on the level of sets is defined as the constant function. This function is not injective. Then I tried to see if this morphism is monic. So let $\gamma: X\to F(X)$ be another object of $\mathscr C$ and consider two morphisms $\gamma\to \alpha$ whose underlying functions are $x,y:X\to A$. Suppose the morphisms with underlying functions $h\circ x$ and $h\circ y$ are equal. Need to check if it must be the case that $x=y$ or not. The first problem is that I don't know how to do this in the abstract setting where $X$ is just some unknown set. Secondly, I just picked $B$ and $\beta$ at random, but I don't understand what properties should $B$ and $\beta$ possess in order to find what I need.

Answer 1

I think this is false.

Let's go through the details. If I made a mistake in my argument, then perhaps that will show us how to prove it if it does turn out to be true.

Let $F(X)$ be the functor that sends a set $X$ to the subset of $X^3$ consisting of points with at least two coordinates equal.

Let's first answer the question of what a morphism from $(C,\gamma : C\to FC)$ to the object $(A,\alpha)$ of the question looks like.

By definition, a morphism is a map of sets $f:C\to A$ such that $\alpha f = F(f)\gamma$. Given such a morphism $f$, we can let $C_\bullet = f^{-1}(\bullet)$, where $\bullet \in A$, and the commutativity requirement says that $\gamma(C_a)\subseteq C_a\times C_a\times C_b$, $\gamma(C_b)\subseteq C_a\times C_a\times C_a$, and $\gamma(C_c) \subseteq C_a\times C_b\times C_b$. Conversely, if we have a decomposition of $C$ into three sets $C= C_a\sqcup C_b\sqcup C_c$ satisfying the compatibility with $\gamma$ condition, then defining $f$ by $f(C_a)=a$, $f(C_b)=b$, $f(C_c)=c$ gives a morphism from $(C,\gamma)$ to $(A,\alpha)$.

Note that $f$ is determined by any two of the sets, and we also have certain restrictions on the sets.

For instance, $\pi_1\gamma(C) \subseteq C_a$, so if a morphism $f$ exists, and $C\ne \varnothing$, then $C_a\ne \varnothing$, and this also forces $C_b\ne\varnothing$, since $\pi_3\gamma(C_a)\subseteq C_b$.

Since two components of the points are equal, we can say a bit more about the images of the sets under $\gamma$, we must have $\gamma(C_a) \subseteq \Delta(C_a)\times C_b$, where $\Delta$ is the diagonal map. (In other words, $\gamma$ sends things in $C_a$ to points with first two coordinate equal), and $\gamma(C_c)\subseteq C_a\times \Delta(C_b)$.

Now let's think a bit about morphisms $h$ from $(A,\alpha)$ to $(B,\beta)$. Let $h_a,h_b,h_c$ be the images of $a$, $b$, and $c$ under $h$ respectively. The condition that $\beta h = h^3\alpha$ applied to each of $a$, $b$, and $c$ gives us that $\beta h_a = (h_a,h_a,h_b)$, $\beta h_b = (h_a,h_a,h_a)$, and $\beta h_c = (h_a,h_b,h_b)$.

Let's think about the possibilities for $B$ if we want $h$ to fail to be injective on the level of sets, and yet still a monomorphism. Note that we can assume without loss of generality that $h$ is surjective on the level of sets, otherwise we can replace $B$ by the image of $h$, and the commutativity condition on $\beta$ forces $\beta$ to restrict to the image of $h$, and the natural inclusion of the image of $h$ into $B$ is a monomorphism, so $h$ was a monomorphism if and only if $h$ restricted to its image is a monomorphism.

Now, if $h$ identifies all of the points, $h_a=h_b=h_c$, $B=\{*\}$ is the terminal object, and $(A,\alpha)\to (B,\beta)$ will not be a monomorphism, if we can find any two parallel but not equal maps to $(A,\alpha)$. Let's produce such a pair.

Let $C=\{a,b,d,e\}$, with $\gamma(a) = (a,a,b)$, $\gamma(b)= (a,a,a)$, $\gamma(d) = (a,a,e)$, and $\gamma(e) = (a,a,d)$. The idea here is that I'm thinking of $a$ and $b$ as permanent members of $C_a$ and $C_b$, but $d$ should be able to switch back and forth between the two, and I don't need anything to map to $C_c$.

Then we can define two parallel maps $(C,\gamma)\to (A,\alpha)$ by $f_1(a)=f_2(a)=a$, $f_1(b)=f_2(b)=b$, $f_1(d)=f_2(e)=a$, and $f_2(d)=f_1(e)=b$, and you can check that these are in fact well defined maps.

Now let's suppose that $h$ identifies two points of $A$. If $h_a\ne h_b$, and $h_c=h_a$ or $h_c=h_b$, then we run into an impossibility, since we would have to have $\beta h_c=\beta h_a$ or $\beta h_c=\beta h_b$, which doesn't happen in this case, since $h_a\ne h_b$.

So instead we must have $h_a=h_b$. Let's take $B=\{0,1\}$, $\beta(0)=\beta(1)=(0,0,0)$, and $h:A\to B$ defined by $h(a)=h(b)=0$, $h(c)=1$.

But if we apply $h$ to our maps $f_1$ and $f_2$ from before, we see that $h\circ f_1 = h\circ f_2$ are both maps sending all of $a,b,d,e$ to $0$. Thus this $h$ is not a monomorphism either.

We appear to have exhausted all of the possibilities for $h$ to be a monomorphism without being injective.

Perhaps someone will spot an error in this though.

Edit notes

In light of Alex Kruckman's excellent answer (+1), it seems very likely that the question has a typo, and is false as written, so the logic in my answer is probably correct.

Answer 2

Just to fix some terminology (and motivation): Let $F\colon \mathscr{A}\to \mathscr{A}$ be a functor. A pair $(A,\alpha)$, where $A\in \mathscr{A}$ and $\alpha$ is an arrow $A\to F(A)$, is called an $F$-coalgebra. If $(A,\alpha)$ and $(B,\beta)$ are $F$-coalgebras, an arrow $h\colon A\to B$ such that $F(h)\circ \alpha = \beta\circ h$ is called a morphism of $F$-coalgebras. So the category $\mathscr{C}$ defined in the question is the category of $F$-coalgebras. There is an obvious forgetful functor $U\colon \mathscr{C}\to \mathscr{A}$, and you are looking for an example of a monic morphism $h$ in $\mathscr{C}$ such that $U(h)$ is not monic in $\mathscr{A}$.

jgon's answer correctly shows that the question as written does not have a positive answer. I suspect there is a typo in the question - if we adjust the example so that $\alpha(b) = (a,\mathbf{b},a)$, then the question admits a positive answer.

To be precise, $A = \{a,b,c\}$ and $\alpha\colon A\to F(A)$ is defined by \begin{align*} \alpha(a) &= (a,b,b)\\ \alpha(b) &= (a,b,a)\\ \alpha(c) &= (a,a,b) \end{align*}

Now the point is that for any $F$-coalgebra $(C,\gamma)$, there is at most one $F$-coalgebra morphism $h\colon (C,\gamma)\to (A,\alpha)$. Indeed, suppose $h\colon (C,\gamma)\to (A,\alpha)$ is an $F$-coalgebra morphism. Let $x\in C$, and let $\gamma(x) = (y_1,y_2,y_3)$. By definition of $F$, at least two of $y_1,y_2,y_3$ are equal. Suppose $y_1 = y_2 = y$. Then $\alpha(h(x)) = F(h)(\gamma(x)) = F(h)(y,y,y_3) = (h(y),h(y),h(y_3))$. The first two coordinates of this triple are equal, so we must have $h(x) = c$ (since the first two coordinates of $\alpha(a)$ and $\alpha(b)$ are not equal). Similarly, if $y_1 = y_3$, then we must have $h(x) = b$, and if $y_2 = y_3$, then we must have $h(x) = a$.

It follows that any $F$-coalgebra morphism $m\colon (A,\alpha)\to (B,\beta)$ is monic, since if $h_1,h_2\colon (C,\gamma)\to (A,\alpha)$ are two $F$-coalgebra morphisms such that $m\circ h_1 = m\circ h_2$, then $h_1 = h_2$ (simply because there is at most one $F$-coalgebra morphism $(C,\gamma)\to (A,\alpha)$).

In your question, you correctly note that we can take $(B,\beta)$ to be the terminal $F$-coalgebra. Then the unique $F$-coalgebra morphism $m\colon (A,\alpha)\to (B,\beta)$ is monic but not injective on underlying sets.