I just read the definition of a coalgebra, defined categorically ( reversing the arrows of the Algebra category ), the given example in the text I am referring to is the Homology on a topological space. I am not familiar yet with algebraic topology, I barely recall something about homotopy groups.

So I am trying to come up with simpler examples, hence the questions:
What is an elementary example of a Coalgebra ? (if any)
How would you define it without using the language of Category ? i.e. for a set $X$ what are the axioms needed to tell that $(X,+,\times)$ has a Coalgebra structure ?
well here $(X,+)$ must be a $\mathbb{K}$-module as in an Algebra, but what to say about the comultiplication $\times$ ?
( and why do we call it co- , and why it is the dual of the multiplication )

This push me to think about how to transform any Categorical definition to the axiomatic definition...

  • $\begingroup$ Do you mean a coalgebra of a monad or a coalgebra of a functor? $\endgroup$
    – N. Virgo
    Commented Jan 8, 2022 at 3:10
  • $\begingroup$ Interesting, so there are still other things that share the same name... I asked about the K-modules that have an associative map $C\rightarrow C\otimes C$ $\endgroup$
    – NotaChoice
    Commented Jan 8, 2022 at 3:19
  • $\begingroup$ should I add which behaves well with respect to scalar multiplication and distributes over the addition, if that makes sense $\endgroup$
    – NotaChoice
    Commented Jan 8, 2022 at 3:20
  • $\begingroup$ Ah, this is outside my knowledge, then. (Probably the thing you're asking about is a special case of one of the things I said, but I don't know.) $\endgroup$
    – N. Virgo
    Commented Jan 8, 2022 at 3:33
  • $\begingroup$ @NotaChoice: If I recall correctly, though they were first understood in algebraic topology context, these guys and Hopf algebras, naturally appear in combinatorics. Try googling for them. $\endgroup$
    – Bumblebee
    Commented Jan 8, 2022 at 17:28

2 Answers 2


Let's start first with the (de-categorified) definition.

If $k$ is your favorite field, then a $(k)$-coalgebra is

  • A $k$-vector space $A$
  • A $k$-linear map $\Delta : A \to A \otimes A$ (called "comultiplication")
  • A $k$-linear map $\epsilon : A \to K$ (called the "counit")

so that

  • $(\Delta \otimes \text{id}_A) \circ \Delta = (\text{id}_A \otimes \Delta) \circ \Delta$ (the "coassociativity" axiom)
  • $(\epsilon \otimes \text{id}_A) \circ \Delta = \text{id}_A = (\text{id}_A \otimes \epsilon) \circ \Delta$ (the "coidentity" axiom)

Here, as usual $\otimes$ is the tensor product, and $\text{id}_A$ is the identity map on $A$.

Ok, this is a weird looking structure. Why should we care about it? Well recall the usual definition of a $(k)$-algebra. It's

  • A $k$-vector space $A$
  • A $k$-bilinear map $m : A \times A \to A$ (called "multiplication")
  • A $k$-linear map $1 : k \to A$ (called the "unit")

so that

  • $m \circ (m \times \text{id}_A) = m \circ (\text{id}_A \times m)$ (the "associativity" axiom)
  • $m \circ (\text{id}_A \times 1) = \text{id}_A = m \circ (1 \times \text{id}_A)$ (the "identity" axiom)

Now if you stare at these requirements for a while, you'll find that they're perfectly backwards. Where algebras have a map $A \times A \to A$, coalgebras have a map $A \to A \otimes A$. Where associativity says that the two obvious ways to multiply three things $A \times A \times A \to A$ are actually the same, coassociativity says that the two obvious ways to comultiply into three things $A \to A \otimes A \otimes A$ are the same.

This is where the "co" prefix comes from. We use "co" in category theory to indicate that we are doing something "with the arrows reversed". Since these axioms come from taking the axioms of an algebra and "turning the arrows around", we call the resulting structure a "co-algebra".

If you're thinking "why would anyone care about these things?", you're not alone. A priori the definition doesn't look particularly natural, and honestly I don't know of any reasons to care about just coalgebras (at least at time of writing). But I do know why we might care about Hopf Algebras. A hopf algebra is a vector space that is both an algebra and a coalgebra in a way that the two structures are "compatible" (this is mediated by an "antipode" $s$). Hopf algebras have use in representation theory, physics, and combinatorics, and you can find some reasons to care about them (and, by extension, coalgebras) here. There are also some discussions here about examples of coalgebras (and, by extension, why we might care).

Here's probably the simplest example of a coalgebra:

Let $A = k^2$, with basis elements $x$ and $y$. Then let

  • $\Delta(x) = x \otimes x$ and $\Delta(y) = y \otimes y$
  • $\epsilon(x) = \epsilon(y) = 1$

where we extend these maps to all of $A$ by linearity.

Then notice

$$ x \overset{\Delta}{\longmapsto} x \otimes x \overset{\Delta \otimes \text{id}_A}{\longmapsto} (x \otimes x) \otimes x $$


$$ x \overset{\Delta}{\longmapsto} x \otimes x \overset{\text{id}_A \otimes \Delta}{\longmapsto} x \otimes (x \otimes x) $$

are actually the same map (likewise when applied to $y$). This is the coassociativity axiom. I'll leave it as an instructive exercise to check the coidentity axiom.

As a slightly less trivial example, consider $k[x]$, which has a basis $\{x^0, x^1, x^2, x^3, \ldots\}$. Then we get a coalgebra structure by considering

  • $\Delta(x^n) = \sum_k x^k \otimes x^{n-k}$
  • $\epsilon(x^0) = 1$
  • $\epsilon(x^n) = 0$ for $n > 1$

again, extended linearly.

Now it's a bit more tedious to check the axioms, but it's doable if you persevere.

This coalgebra apparently has applications in combinatorics, though I admit I don't know much about this myself. For more you might look into Joni and Rota's Coalgebras and Bialgebras in Combinatorics, available here, say.

I hope this helps ^_^

  • 1
    $\begingroup$ I always thought of it kind of like, in the natural numbers then we have a product $m*n$, and we could make a coproduct which would for example have $\Delta(10) = 1\otimes 10 + 2\otimes 5$, and similarly for other integers it enumerates all of their possible factorisations, and then it would be the extension of this but to an algebra as you've defined. Does that make sense? $\endgroup$
    – Jojo
    Commented Jan 8, 2022 at 12:35
  • $\begingroup$ Thank you, I will check the links, I take it that Algebras in general need not be unital (have a unit) but Coalgebras need it always? $\endgroup$
    – NotaChoice
    Commented Jan 8, 2022 at 16:15
  • $\begingroup$ @Joe -- that definately makes sense, and the intuition that the comultiplication (I don't like the word "coproduct" because it overloads the categorical notion) sums over all the ways to "decompose" an element is a really good one! In fact, this is the view espoused in the Joni and Rota paper I linked, and is responsible for the connections with combinatorics (where counting the ways to decompose an object is a very common problem) $\endgroup$ Commented Jan 8, 2022 at 19:01
  • 1
    $\begingroup$ @NotaChoice -- I'm most familiar with counital coalgebras, which is why I included the counit in my definition. There are people who study non-counital coalgebras (analogous to nonunital algebras), though I've never really looked into them. See here for some discussion. $\endgroup$ Commented Jan 8, 2022 at 19:04
  • 4
    $\begingroup$ would the downvoter please explain their issue with this answer? $\endgroup$ Commented Jan 11, 2022 at 19:44

The simplest examples arise from free modules with chosen bases (over a commutative ring; for a non-commutative ring you have to use bimodules to get the tensor product to work).

Explicitly, the tensor product $M\otimes M$ of a a free $R$-module $M$ with chosen basis $X\subseteq M$ has basis $\{x_1\otimes x_2:x_1,x_2\in X\}\cong X\times X$. It turns out that the choice of basis $X$ for $M$ gives a coalgebra structure with comultiplication the linear map $M\to M\otimes M$ generated by $x\mapsto x\otimes x$ and that has a counit the linear map $M\to R$ generated by $x\mapsto 1$. More explcitly, the comultiplication is given by $\sum_ir_ix_i\mapsto\sum r_ix_i\otimes x_i$, and the counit by $\sum_i r_ix_i\mapsto\sum_ir_i$.

The next simplest examples arise from free modules of finite rank (i.e. admitting a finite basis) with an algebra structure. Namely, given an $R$-algebra $M$ with multiplication $m\colon M\otimes M\to M$ and unit $e\colon R\to M$, these correspond to an $R$-linear map $m^\vee\colon M^\vee\to(M\otimes M)^\vee$ and an $R$-linear map $e^\vee\colon M^\vee\to R$, where $M^\vee=\{f\colon M\to R\colon f$ is $R$-linear$\}$. In the case $M$ is free of finite rank, we have an isomorphism $(M\otimes M)^\vee\cong M^\vee\otimes M^\vee$, so a (unital) algebra structure on $M$ induces a coalgebra (with counit) structure on $M^\vee$, the $R$-module dual to $M$.


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