Externalising monoids using the Yoneda-embedding and relation to Kleisli categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:41:16Z http://mathoverflow.net/feeds/question/103519 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103519/externalising-monoids-using-the-yoneda-embedding-and-relation-to-kleisli-categori Externalising monoids using the Yoneda-embedding and relation to Kleisli categories Kris Joanidis 2012-07-30T15:41:34Z 2012-08-01T21:47:31Z <p>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.</p> <p>Step 1. <em>Externalise</em> the monoid by turning it into a monoid $(Hom(I,X),\cdot,\bot)$ where for $a,b : I \rightarrow X$ define $a \cdot b = \triangledown \circ (a \otimes b) \circ \lambda_I ^{-1}$.</p> <p>Step 2. <em>Represent</em> every element of $Hom(I,X)$ as and element of $Hom(X,X)$ by defining</p> <p>$Y : Hom(I,X) \rightarrow Hom(X,X)$</p> <p>$Y a = \triangledown \circ (a \otimes id_x) \circ \lambda_X ^{-1}$</p> <p>(where $\lambda_X : X \otimes I \rightarrow X$ is the appropriate structure map)</p> <p>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 $(a\cdot)$.</p> <p>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?</p> <p>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 <a href="http://en.wikipedia.org/wiki/Kleisli_category#Formal_definition" rel="nofollow">definition of the Kleisli category</a> and Step 2 looks like the $^\star$ operator used in the <a href="http://en.wikipedia.org/wiki/Kleisli_category#Extension_operators_and_Kleisli_triples" rel="nofollow">definition of Kleisli triples</a>. Are these constructions instances of a more general one?</p> <p><a href="http://arxiv.org/abs/0812.2266" rel="nofollow">Here's a link to the paper</a></p> http://mathoverflow.net/questions/103519/externalising-monoids-using-the-yoneda-embedding-and-relation-to-kleisli-categori/103524#103524 Answer by Zhen Lin for Externalising monoids using the Yoneda-embedding and relation to Kleisli categories Zhen Lin 2012-07-30T16:37:56Z 2012-07-31T00:41:57Z <p>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 <em>any</em> object $X$ in $\mathcal{C}$, the hom-set $\mathcal{C}(X, M)$ is an ordinary monoid in $\textbf{Set}$.</p> <p>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 $\mathcal{C}(X, M)$.</p> <p>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.</p> <p>But hope is not lost yet: it turns out $\mathcal{C}(I, -)$ <em>is</em> 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.</p> <p>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 $I_\mathcal{D} \to F I_\mathcal{C}$.</p> <p>More generally, $\mathcal{C}(X, -)$ is lax monoidal whenever $X$ is a <em>comonoid</em>. This recovers the result of the first paragraph, since in a cartesian monoidal category, <em>every</em> object is a comonoid in a unique way.</p> <hr> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/103519/externalising-monoids-using-the-yoneda-embedding-and-relation-to-kleisli-categori/103559#103559 Answer by Michal R. Przybylek for Externalising monoids using the Yoneda-embedding and relation to Kleisli categories Michal R. Przybylek 2012-07-30T22:51:23Z 2012-08-01T21:47:31Z <p>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 $U \colon \mathbb{C}$-$\mathbf{Cat} \rightarrow \mathbf{Cat}$.</p> <p>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 $1_X$.</p> <p>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 $[X, X] \in \mathbb{C}$.</p> <p>We should get the second construction by applying the underlying functor to $X \rightarrow [X, X]$.</p> <hr> <p>I will try to elaborate a bit more on the subject.</p> <p>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</p> <p>$$\mathit{nat}(\hom(-, 1), \hom(-, 1))$$ 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 $\mathbb{C}$.</p> <p>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 $E \colon 1_X \rightarrow 1_{[X, X]}$.</p> <p>The second construction is given by the application of the underlying functor $U$ to $E$:</p> <p>$$U(E) \colon U(1_X) \rightarrow U(1_{[X, X]})$$</p> <p>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.</p> <p>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:</p> <p>$$F \otimes_\mathbb{C} G = \int^{B, C} F(B) \times G(C) \times \hom(-, B \otimes C)$$ </p> <p>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 $\mathbf{Set}^{\mathbb{C}^{op}}$-$\mathbf{Cat}$.</p> <p>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:</p> <p>$$V(E) \colon U(1_X) = V(1_{y_\mathbb{C}(X)}) \rightarrow V(1_{[y_\mathbb{C}(X), y_\mathbb{C}(X)]})$$</p>