Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: path objects of bicats

Daniel Plácido (Jun 24 2021 at 22:28):

The following excerpt is from Lack's "A Quillen model structure for bicategories", where $\mathcal B$ is a bicategory.
image.png

I'm having a hard time understanding what the 2-functors $D$ and $(P,Q)$ are doing to the 2-cells. A 2-cell in $\mathcal{PB}$ looks like this:
image.png

Ok, so for instance $D$ is sending $b$ to the vertical-arrow-identity, and a morphism $f:x\to y$ to the identity of $f$:
image.png

There's really no place to send a 2-cell $\alpha:f\Rightarrow f'$ in this case. The same issue happens to $P$ and $Q$.

Did I miss something? If not, perhabs Lack meant for $\gamma,\gamma'$ to be 2-cells between the horizontal arrows in the morphism of $\mathcal{PB}$? (edit: possibly not, the vertical arrows being equivalences implies we require 2-cells to be horizontally directed)

Amar Hadzihasanovic (Jun 25 2021 at 09:06):

In your first picture, if the 2-cell is from $\alpha$ to $\psi'$, then $\beta_1$ actually goes the other way (from $f$ to $f'$) and in the equation you have to attach it to the left of $\psi'$.

Amar Hadzihasanovic (Jun 25 2021 at 09:07):

So it's like $\alpha \beta_2 = \beta_1 \psi'$.

Amar Hadzihasanovic (Jun 25 2021 at 09:08):

And if you use the same convention as the diagram above, $f$ is sent by $D$ to a diagram rotated wrt the one you drew: the “horizontal” arrows are identities and the “vertical” ones are $f$. And then you should be able to see that there is, in fact, a place to send a 2-cell $\alpha: f \Rightarrow f'$.

Daniel Plácido (Jun 25 2021 at 22:57):

Thanks, Amar, this really helped, the $\alpha\beta_2 = \beta_1\psi'$ makes much sense. I was screwing things up because my $f$ is actually Lack's $b$, and Lack uses $f$ for the other direction...

Daniel Plácido (Jun 25 2021 at 23:00):

$D$ ends up looking like this: image.png (the objects of $\mathcal{PB}$ are the vertical arrows in the squares)

Daniel Plácido (Jun 25 2021 at 23:05):

But how is $D$ a biequivalence? For instance let's try to check for biessential surjectivity, i.e. every $d\in\mathcal{PB}$ has an equivalence to $Fc$, where $c\in\mathcal B$. An equivalence in $\mathcal{PB}$ looks like this:
image.png

Issue here being: if $d$ and $c$ were equivalences, we would have enough the data for an equivalence to $Dy$, but in fact only $f$ and $g$ are required to be equivalences, not the vertical arrows. Again I'm in a pit here.

Daniel Plácido (Jun 25 2021 at 23:05):

Perhabs we have to ask for all arrows in the squares to be equivalences?

Daniel Plácido (Jul 01 2021 at 16:47):

I managed to go through this, saying just to "get of out of queue"