Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm reading a paper in which the following construction appears: Let $(\mathcal{C}, \otimes, I)$ be a (dagger) monoidal category, and $(X, \triangledown, \bot)$ be a monoid in this category.

Step 1. Externalise the monoid by turning it into a monoid $(Hom(I,X),\cdot,\bot)$ where for $a,b : I \rightarrow X$ define $a \cdot b = \triangledown \circ (a \otimes b) \circ \lambda_I ^{-1}$.

Step 2. Represent every element of $Hom(I,X)$ as and element of $Hom(X,X)$ by defining

$Y : Hom(I,X) \rightarrow Hom(X,X)$

$Y a = \triangledown \circ (a \otimes id_x) \circ \lambda_X ^{-1}$

(where $\lambda_X : X \otimes I \rightarrow X$ is the appropriate structure map)

The author claims this is an example of the Yoneda-embedding. I can almost see how this is the case as $Y a$ is similar to $(a\cdot)$.

My questions are essentially: is this a standard construction, where can I read more about it? Is it really an instance of the Yoneda-embedding?

Also the two steps seem to be similar to the construction of the Kleisli category of a monad. The Wikipedia article on the Kleisli category includes two different presentations. Step 1 resembles the definition of the Kleisli category and Step 2 looks like the $^\star$ operator used in the definition of Kleisli triples. Are these constructions instances of a more general one?

Here's a link to the paper

share|improve this question

2 Answers 2

up vote 4 down vote accepted

For any monoidal category $\mathbb{C}$ there exists the "underlying" monoidal functor $\hom(I, -) \colon \mathbb{C} \rightarrow \mathbf{Set}$. As is the idea of a monoidal functor, it preserves structures defined by monoidal operations. Particularly, $\hom(I, -)$ lifts to the "underlying" 2-functor from the 2-category of $\mathbb{C}$-enriched categories to the 2-category of ordinary (that is: $\mathbf{Set}$-enriched) categories $U \colon \mathbb{C}$-$\mathbf{Cat} \rightarrow \mathbf{Cat}$.

A monoid $X$ internal to $\mathbb{C}$ is precisely a $\mathbb{C}$-enriched category $1_X$ having a single object $1$ and $\hom(1, 1) = X$. From this perspective, the first construction corresponds to taking the underlying category of $1_X$.

If $\mathbb{C}$ is closed and has equalisers, then something much stronger then proposition 2.6 should be true (i.e. proposition 2.6 should hold internally to $\mathbb{C}$). The monoid $1_X$ via Yoneda $y_{1_X} \colon 1_X \rightarrow \mathbb{C}^{1_X^{op}}$ embeds into the category of presheaves on $1_X$. Now the enriched Yoneda lemma says that $X$ is isomorphic to the object of natural transformations $\mathit{nat}(\hom(-, 1), \hom(-, 1))$, which, by the definition of a natural transformation, is a regular subobject of $[X, X] \in \mathbb{C}$.

We should get the second construction by applying the underlying functor to $X \rightarrow [X, X]$.

I will try to elaborate a bit more on the subject.

Let us assume that $\mathbb{C}$ is symmetric monoidal closed and has equalisers. Then any monoid $X$ internal to $\mathbb{C}$ admits embedding $y_{1_X} \colon 1_X \rightarrow \mathbb{C}^{1_X^{op}}$. The object of natural transformations

$$\mathit{nat}(\hom(-, 1), \hom(-, 1))$$ by definition is the equaliser of $l, r \colon [X, X] \rightarrow [X, [X, X]]$ in $\mathbb{C}$, where $l$ is the transposition of: $$\mu_\mathbb{C} \circ (\mathit{id}_{[X, X]} \otimes \nabla^*) \colon [X, X] \otimes X \rightarrow [X, X]$$ $r$ is the other transposition of: $$\mu_\mathbb{C} \circ (\nabla^* \otimes \mathit{id}_{[X, X]}) \colon X \otimes [X, X] \rightarrow [X, X]$$ $\nabla^* \colon X \rightarrow [X, X]$ is the transposition of the monoidal multiplication $X \otimes X \rightarrow X$, and $\mu_\mathbb{C} \colon [X, X] \otimes [X, X] \rightarrow [X, X]$ is the internalised composition from $\mathbb{C}$.

The enriched Yoneda lemma says that $$\mathit{nat}(\hom(-, 1), \hom(-, 1)) \approx \hom(1, 1) = X$$ Therefore the "arrows part" of $hom(-, 1)$ --- $e \colon X \rightarrow [X, X]$ is the equaliser of $l, r \colon [X, X] \rightarrow [X, [X, X]]$. Furthermore, because $hom(-, 1)$ is a functor, it maps the composition in $1_X^{op}$ to the composition in $\mathbb{C}$, turning $e$ into a functor between internal monoids $E \colon 1_X \rightarrow 1_{[X, X]}$.

The second construction is given by the application of the underlying functor $U$ to $E$:

$$U(E) \colon U(1_X) \rightarrow U(1_{[X, X]})$$

One may perhaps use the weak version of the Yoneda lemma to construct $U(E)$ in case $\mathbb{C}$ is not monoidal closed with equalisers. However, there is also a more natural solution.

Let us recall that if $\mathbb{C}$ is monoidal, then its category of presheaves $\mathbf{Set}^{\mathbb{C}^{op}}$ inherits the monoidal structure via the very special case of convolution:

$$F \otimes_\mathbb{C} G = \int^{B, C} F(B) \times G(C) \times \hom(-, B \otimes C)$$

Moreover, Brian Day showed that $\otimes_\mathbb{C}$ makes $\mathbf{Set}^{\mathbb{C}^{op}}$ a monoidal (bi)closed category, with the Yoneda embedding $y_\mathbb{C} \colon \mathbb{C} \rightarrow \mathbf{Set}^{\mathbb{C}^{op}}$ preserving the structure (i.e. not only does $y_\mathbb{C}$ preserve tensors, but any existing linear exponents). This means that $y_\mathbb{C}$ rises to the 2-functor $Y \colon \mathbb{C}$-$\mathbf{Cat} \rightarrow \mathbf{Set}^{\mathbb{C}^{op}}$-$\mathbf{Cat}$. By Yoneda, the underlying functor $U \colon \mathbb{C}$-$\mathbf{Cat} \rightarrow \mathbf{Cat}$ factors through $Y$ followed by the underlying functor $V$ of $\mathbf{Set}^{\mathbb{C}^{op}}$-$\mathbf{Cat}$.

Since the Yoneda functor $y_\mathbb{C}$ also preserves equalisers, every monoid $X$ in $\mathbb{C}$ has a representation as a submonoid of $y_\mathbb{C}(X)^{y_\mathbb{C}(X)}$ in $\mathbf{Set}^{\mathbb{C}^{op}}$, and $X$ is a submonoid of $[X, X] \in \mathbb{C}$ iff the linear exponent $[X, X]$ exists in $\mathbb{C}$. "The second construction" is:

$$V(E) \colon U(1_X) = V(1_{y_\mathbb{C}(X)}) \rightarrow V(1_{[y_\mathbb{C}(X), y_\mathbb{C}(X)]})$$

share|improve this answer

Let's start with the cartesian monoidal case, since it is the easiest one to understand. Recall that the Yoneda embedding is left exact, so if we have a monoid $M$ in a category $\mathcal{C}$, then $\mathcal{C}(-, M)$ is automatically a monoid in the presheaf topos $[\mathcal{C}^\textrm{op}, \textbf{Set}]$, and for any object $X$ in $\mathcal{C}$, the hom-set $\mathcal{C}(X, M)$ is an ordinary monoid in $\textbf{Set}$.

More explicitly, if the structural data of $M$ are $e : 1 \to M$ and $m : M \times M \to M$, we get induced maps $e_* : \mathcal{C}(X, 1) \to \mathcal{C}(X, M)$ and $m_* : \mathcal{C}(X, M \times M) \to \mathcal{C}(X, M)$; but left exactness means that $\mathcal{C}(X, 1) \cong 1$ and $\mathcal{C}(X, M \times M) \cong \mathcal{C}(X, M) \times \mathcal{C}(X, M)$, so this indeed induces the structure of a monoid on $\mathcal{C}(X, M)$.

Now let $\mathcal{C}$ be a monoidal category. I will pretend it is strict monoidal. Let $M$ be a monoid in $\mathcal{C}$. As before, we get maps $\mathcal{C}(X, I) \to \mathcal{C}(X, M)$ and $\mathcal{C}(X, M \otimes M) \to \mathcal{C}(X, M)$, but unfortunately there is no obvious map $1 \to \mathcal{C}(X, I)$ or $\mathcal{C}(X, M) \times \mathcal{C}(X, M) \to \mathcal{C}(X, M \otimes M)$; that is to say, $\mathcal{C}(X, -)$ is not automatically a lax monoidal functor.

But hope is not lost yet: it turns out $\mathcal{C}(I, -)$ is a lax monoidal functor! There is an obvious map $1 \to \mathcal{C}(I, I)$ (namely the constant map with value $\textrm{id}_I$) and a natural map $\mathcal{C}(I, Y) \times \mathcal{C}(I, Z) \to \mathcal{C}(I, Y \otimes Z)$ given by $(f, g) \mapsto f \otimes g$ (and suppressing the isomorphism $I \otimes I \cong I$, as previously mentioned), and it is easy to check that this does indeed give $\mathcal{C}(I, -)$ the structure of a lax monoidal functor.

Finally, it is a well-known fact that lax monoidal functors carry monoids to monoids. Let's prove this now. Suppose $F : \mathcal{C} \to \mathcal{D}$ is a lax monoidal functor and $M$ is a monoid in $\mathcal{C}$. Then we get a morphism $I_\mathcal{D} \to F M$ by composing $I_\mathcal{D} \to F I_\mathcal{C}$ with $F e : F I_\mathcal{C} \to F M$, and a morphism $F M \otimes_\mathcal{D} F M \to F M$ by composing $F M \otimes_\mathcal{D} F M \to F (M \otimes_\mathcal{C} M)$ with $F m : F (M \otimes_\mathcal{C} M) \to F M$. The coherence axioms for lax monoidal functors imply that $F M$ together with these structural data satisfy the monoid axioms. For example, to check that the right unit axiom holds, we must show that the composite $$F M \to F M \otimes_\mathcal{D} I_\mathcal{D} \to F M \otimes_\mathcal{D} F M \to F (M \otimes_\mathcal{C} M ) \to F M$$ is equal to the identity, but this is equal to the composite $$F M \to F M \otimes_\mathcal{D} I_\mathcal{D} \to F M \otimes_\mathcal{D} F I_\mathcal{C} \to F (M \otimes_\mathcal{C} I_\mathcal{C}) \to F (M \otimes_\mathcal{C} M ) \to F M$$ by naturality of $F Y \otimes_\mathcal{D} F Z \to F (Y \otimes_\mathcal{C} Z)$, and this is equal to the identity by the right unit axiom for $M$ together with the coherence axiom for $I_\mathcal{D} \to F I_\mathcal{C}$.

More generally, $\mathcal{C}(X, -)$ is lax monoidal whenever $X$ is a comonoid. This recovers the result of the first paragraph, since in a cartesian monoidal category, every object is a comonoid in a unique way.

Step 2 of your question is also related to the Yoneda embedding, but in a more subtle way. In effect, it is externalising the internal monoid $M$ by considering its left self-action. This is basically an appeal to the monoid version of Cayley's theorem, which is sometimes considered a special case of the Yoneda embedding.

Personally, I don't see a connection with the Kleisli category construction. Perhaps one can regard the "embedding" of $\mathcal{C}(I, X)$ into $\mathcal{C}(X, X)$ as an instance of a "funny composition", but I don't think there's anything deeper than that.

share|improve this answer
Thank you for your answer. I will take time to go through it in detail. Could you please clarify the answer you gave about step 2? I was considering this, by turning the set $\mathcal{C}(I,X)$ into a category with one object and using the Yoneda-embedding, but that gives a monoid on $\mathcal{Set}(\mathcal{C}(I,X),\mathcal{C}(I,X))$ rather than $\mathcal{C}(X,X)$. Can this be fixed? –  Kris Joanidis Jul 31 '12 at 9:27
I was deliberately imprecise. A more accurate version would be to say that the external monoid $\mathcal{C}(I, M)$ acts on $M$ by the given formula – which can be regarded as the formula for the left self-action of $M$ "parametrised" by $I$. Alternatively, with more assumptions on $\mathcal{C}$ one can use a more elegant construction as in Michal's answer – what is happening belongs more in the realm of enriched category theory. –  Zhen Lin Jul 31 '12 at 10:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.