6
$\begingroup$

I'd like to know a reference where the following property, that I believe to be true, is checked: given a diagram of categories and functors $B\leftarrow A\rightarrow C$ (I'm actually interested in the enriched case, over a closed symmetric monoidal category), if $A\rightarrow C$ is fully faithful then so is the functor to the (strict) push-out $B\rightarrow B\cup_A C$.

I should say that I have been able to check this in some specific cases that I needed, but a reference would help me to reduce a paper in four pages ;-)

$\endgroup$
4
  • $\begingroup$ @Adeel, thanks, I knew that paper, it says that for ordinary categories $B\r B\cup_AC$ is an inclusion. That's a hint, in my opinion, about the fact that my guess is true. I'd like it to be also full, and for enriched categories. $\endgroup$ Commented Jan 22, 2014 at 17:39
  • 1
    $\begingroup$ On the other hand, the proofs in the cited paper heavily use the explicit combinatorial description of morphisms in the pushout, so it seems unlikely to me that this will generalise to the enriched case. $\endgroup$
    – Zhen Lin
    Commented Jan 23, 2014 at 9:53
  • $\begingroup$ @ZhenLin, I think it does, everything can be done diagramatically, via tensor products and push-outs. I hope somebody has done something like that before! Or maybe there is a short argument that I cannot find. $\endgroup$ Commented Jan 23, 2014 at 10:12
  • 1
    $\begingroup$ You may be interested in contributing to a proposal Spanish language version of math stackexchange; it could use some input from fluent professors: area51.stackexchange.com/proposals/64529/… $\endgroup$ Commented Feb 1, 2014 at 3:13

1 Answer 1

8
$\begingroup$

A proof of this fact is given in Proposition 3.1 of

Alexandru Stanculescu, Constructing model categories with prescribed fibrant objects, arXiv:1208.6005.

He attributes this result to an earlier paper from the 80's by Fritsch and Latch.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! It's incredible I didn't notice that result in Stanculescu's paper, that I had been looking at. $\endgroup$ Commented Jan 25, 2014 at 1:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .