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

Let $P: \mathcal{C}\to \mathcal{A}$ be a functor. We call a morphism $v$ of $\mathcal{C}$ vertical (over $A\in \mathcal{A}$) if $P(v)=1_A$; then we have the fibre category $\mathcal{C}_A$ of vertical morphisms over $A$. We call a morphism $f: X\to Y$ in $\mathcal{C}$ cartesian if for $f': X'\to Y$ with $P(f)=P(f')$, there exists a unique vertical morphism $v: X'\to X$ such that $f'=f\circ v$.

If $f: A\to B$ is in $\mathcal{A}$ and $Y\in \mathcal{C}(B)$, we call a cartesian morphism of type $\theta_f(Y): f^\ast Y\to Y$ with $P(\theta_f)=f)$ a cartesian $f$-lifting of $Y$. If there exists such for each $f$ and $Y$, we call $P$ a prefibration, and a precleavage is a choice of $\theta_f(Y)$ for each $f$ and $Y$. If we choose $\theta_{1_A}=1_A$, the preclivage is called normal.

After fixing a cleavage, we obtain a functor $\theta_f: \mathcal{C}(B)\to \mathcal{C}(A)$, and for $X\xrightarrow{f}Y\xrightarrow{g}Z$ there is a transformation $c_{g, f}: f^\ast\circ g^\ast\Rightarrow (g\circ f)^\ast$ (considering $ \theta_g(Z)\circ \theta_f(g^\ast(Z))$). Thus, we have a lax functor $\textbf{C}: \mathcal{A}^{op}\to CAT$ defined as $\textbf{C}(A)=\mathcal{C}(A)\ A\in\mathcal{A}$, $\textbf{C}(f)=f^{\ast}$,

$C_{g,f}=c_{g, f}: \textbf{C}(f)\circ \textbf{C}(g)\Rightarrow \textbf{C}(g\circ f)$.

Let $P: \mathcal{C}\to \mathcal{A}$ and $Q: \mathcal{D}\to \mathcal{A}$ be prefibrations with fixed normal cleavages. If $T: \mathcal{C}\to \mathcal{D}$ is a functor over $\mathcal{A}$, i.e. with $Q\circ T= P$, then by restriction we have functors $T_A: \textbf{C}(A)\to \textbf{D}(A)$. Applying $T$ to $\theta_f(Y)$ and considering the cartesian $\theta_f(T(Y))$ (in $\mathcal{D}$), we have a natural vertical morphism $T_f(Y): T(f^\ast(Y))\to f^\ast(T(Y))$, which defines a transformation $T_f: T_A\circ \textbf{C}(f)\Rightarrow \textbf{D}(f)\circ T_B$. The data $(T_A, T_F)_{A, f}$ is what is called an oplax transformation. In the other direction, from an oplax transformation $(T_A, T_F)_{A, f}$ we can make a functor $T: \mathcal{C}\to \mathcal{D}$ over $\mathcal{A}$ (considering that each morphism over $\mathcal{C}$ or in $\mathcal{D}$ has a unique factorization as a vertical morphism followed by a cartesian morphism).

Thus the oplax transformations are identified with functors $T$ over $\mathcal{A}$.

My question is:

If (instead of an oplax tranformation) we consider a lax tranformation $T$, i.e. a family

$T_A: \textbf{C}(A)\to \textbf{D}(A)$ and $T_f: \textbf{D}(f)\circ T_B \Rightarrow T_A\circ \textbf{C}(f)$

with the coherence conditions:

$T_{gf}\ast (d_{g, f})\circ T_C)=$

$(T_A\circ c_{g, f})\ast(T_f\circ \textbf{C}(g))\ast(\textbf{D}(f)\circ T_g): \textbf{D}(f)\circ\textbf{D}(g)\circ T_C\Rightarrow T_A\circ \textbf{D}(f)\circ\textbf{D}(g)$

(the unitary condition being unnecessary, by our choice of normal cleavages).

Is there some way to represent $T$ as a categorical construction?

share|improve this question
    
If I understand correctly what you are asking, this is a good question. But before I edited it for spelling, grammar, and punctuation, it was almost impossible to understand. I appreciate that English can be a difficult language for non-native speakers, but as a practical matter, you might have more luck getting answers if your questions were written in a way that was easier to understand. –  Mike Shulman Aug 13 '12 at 6:14
    
And as long as I'm giving advice, it's probably not necessary to rehash the definitions of cartesian arrow, cleavage, and (pre)fibration in the question. Giving background is good, but not so much background that it overwhelms the question being asked -- conciseness is also a virtue, and you can always include a link for more background details. I suppose opinions may differ on this point, though. –  Mike Shulman Aug 13 '12 at 6:17
    
One more thing: if you can be more precise in the question title than "A question about ...", you'll get more people clicking on it. A title for this question might be "Can lax transformations be represented in terms of fibrations?" –  Mike Shulman Aug 13 '12 at 6:19
    
thank you for suggestions –  Buschi Sergio Aug 13 '12 at 16:35

1 Answer 1

I guess that what you mean is, is there some way to represent $T$ in terms of the (pre)fibrations $P:\mathcal{C}\to \mathcal{A}$ and $Q:\mathcal{D}\to \mathcal{A}$? As far as I know, the answer is no — at least, nothing so natural as how oplax transformations are represented by plain functors over $\mathcal{A}$.

Here's a bit more abstract context, to make this answer seem less negative. (-: There's a general construction (due originally to Hermida, I believe -- see his paper From coherent structures to universal properties) which, given a well-behaved 2-monad $M$ on a well-behaved 2-category $\mathcal{X}$, produces a new 2-monad $\hat{M}$ on a new 2-category $\hat{\mathcal{X}}$ such that:

  • $M$-algebras can be identified with $\hat{M}$-algebras;
  • pseudo $M$-morphisms can be identified with pseudo $\hat{M}$-morphisms;
  • colax $M$-morphisms can be identified with colax $\hat{M}$-morphisms; and
  • $\hat{M}$ is colax-idempotent, i.e. every morphism in $\hat{\mathcal{X}}$ between $\hat{M}$-algebras is a colax $\hat{M}$-morphism in a unique way.

One property of colax-idempotent 2-monads is that if a morphism between algebras is lax, then the lax structure map automatically becomes an inverse to the unique colax structure map and makes it pseudo. Thus every lax morphism is automatically pseudo. Hence, in the above theorem we cannot expect to identify lax $\hat{M}$-morphisms with lax $M$-morphisms, unless $M$ already had the very special property that all lax morphisms are pseudo.

For the case in point, $\mathcal{X}=\mathrm{Cat}^{\mathrm{ob}(\mathcal{A})}$ and $M$ is the 2-monad whose algebras are normal lax functors. Then $\hat{\mathcal{X}}$ is $\mathrm{Cat}/\mathcal{A}$ and $\hat{M}$ is the 2-monad whose algebras are prefibrations. Thus, every morphism in $\mathrm{Cat}/\mathcal{A}$ between prefibrations is a colax $\hat{M}$-morphism, corresponding to a colax $M$-morphism, i.e. an oplax transformation — but there is no natural way to see the lax $M$-morphisms in terms of $\hat{M}$-algebras.

Another example (perhaps the other canonical example) is when $\mathcal{X}=\mathrm{Cat}$ and $M$ is the 2-monad whose algebras are monoidal categories. Then $\hat{\mathcal{X}}$ is the 2-category of multicategories, and $\hat{M}$-algebras are representable multicategories.

share|improve this answer

Your Answer

 
discard

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.