TITLE: A monic coalgebra morphism whose underlying $\text{Set}$ morphism is not injective
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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.