Why is subcoalgebra structure unique? Sorry for breaking the harmony of MO with a easy and silly questions. I have stuck on  elementary category-theoretic reasoning about subcoalgebras, namely:
Let $F:\mathcal{Set}\to \mathcal{Set}$ be a functor. An $F$-coalgebra is a pair $\mathcal{A}=(A,\alpha)$ where $\alpha:A\to F(A)$ is an arbitrary map. Given $F$-coalgebras $\mathcal{A}=(A,\alpha)$ and $\mathcal{B}=(B,\beta)$, a homomorphism $\varphi:\mathcal{A}\to \mathcal{B}$ is a map $\varphi:A\to B$ which make the obvious corresponding square commute. 
 
A subset  $U\subseteq A$ is called closed, if an $F$-coalgebra structure $\mathcal{U}=(U,\delta)$ can be defined on $U$ so that the natural inclusion $\subseteq :U\to A$ is a homomorphism. In this case $\mathcal{U}$ is called subcoalgebra of $\mathcal{A}$. 
A coalgebra literature indicates that coalgebra structure map $\delta$ is easily seen to be unique. 


I have stuck on uniqueness, what is the reason, how it can be seen? Is it category-theoretic reason or rather set-theoretic?


 A: Let us say that a monomorphism $U\subseteq A$ is nontrivial if $U$ is nonempty. Since every nontrivial monomorphism in $Set$ is split, every $F : Set\to Set$ takes nontrivial monomorphisms to monomorphisms. In particular, your $F(\subseteq)$ is again a monomorphism in this case.
So if you have two coalgebra structures on $U\subseteq A$ that make the square commute, then they must be equal, since you can cancel $F(\subseteq)$ from the left.
The only case left to treat is the trivial case $U = \emptyset$. But then the coalgebra structure is unique by the initiality.
The uniqueness of the subcoalgebra structure is false in some categories other than $Set$. For example on $Ab$, consider the functor $-\otimes\mathbb{Z}_2$. The quotient map $\mathbb{Z}\to\mathbb{Z}_2$ makes $\mathbb{Z}$ into a coalgebra. Applying the functor to the inclusion of the even integers $\mathbb{Z}\subseteq\mathbb{Z}$ results in the zero map. Therefore the even integers $\mathbb{Z}$ are a subcoalgebra in two different ways: either via the projection or via the zero map $\mathbb{Z}\to\mathbb{Z}_2$.
