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. enter image description here

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?


1 Answer 1


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 $\mathbb{Z}\to\mathbb{Z}_2$.

  • 4
    $\begingroup$ A purist might object by noting that $\emptyset \to S$ can't be split if $S$ is nonempty, but that special case is trivial by initiality of $\emptyset$. $\endgroup$
    – Todd Trimble
    Apr 17, 2018 at 10:50
  • $\begingroup$ @ToddTrimble: of course, thank you for the correction! I'll edit my answer accordingly. $\endgroup$ Apr 17, 2018 at 10:53
  • $\begingroup$ Oh I realized, splitting does a job. Thanks a lot. $\endgroup$ Apr 17, 2018 at 11:41

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