A (p:E---->B)- relative colimit is what you get if you try to make sense of the notion of a colimit of a diagram in E under the constraint that the colimit cocone is sent to some fixed cocone in B.

It's more or less the "universal lift of a cocone". (When the base is $\Delta^0$ all the cocones are equivalent so you instead get the "universal cocone" period).

That's my understanding at least.

If E -> B is a cocartesian fibration where the fibres have I-shaped colimits and the cocartesian pushforwards preserve these, you can describe relative colimits very explicitly: the relative colimit of a diagram I -> E over I^\triangleright -> B (taking the cone point to b) is given by first taking the cocartesian pushforward of the diagram in E to the fibre E_b and then taking the colimit in E_b.

I was wondering about the same issue (how to think about relative (co)limits) the other day and searching for a statement like Rune's in HTT and had trouble finding it. I will try to hunt it down again and add it to the nlab page.

@TimCampion It's the statements around 4.3.1.10 and after if I'm not mistaken

like 4.3.1.10-13 approximately

Yeah, I think that's the one I found, and I thought I must be getting close...

Then 4.3.1.11 is promising, but it doesn't contain the explicit description Rune gave.

I think the proof does though

I suppose it would have to!

The Q and \overline q' in the proof do exactly what Rune describes

I bet Rune could give a proof that I would understand better.

Interesting that right at the beginning of the proof of 4.3.1.11 Lurie just kind of says "take the cocartesian pushforward of the diagram to the cocone fiber" without citing anything saying you can do this. I will have to track down how we know this can be done functorially.

Yeah I want to do a close study of this section.

Isn't that because $Fun(K,X)\to Fun(K,Q)$ is a coCartesian fibration ? So you have a morphism in the latter with domain just the restriction of $f$ to $K$, i.e. $p\circ q$, which you can then lift. The target of the lift will of course be a map with values in $X_s$ because the morphism in $Fun(K,Q)$ had target $s$

So I answered my question that I had already mentioned here earlier (mathoverflow.net/questions/383061/…) and I wrote a sketch of a proof for the 1-categorical case, and isolated at the end the sort of statement I need in the $(\infty,2)$-categorical case to be happy with the result, so if someone knows where I can find this type of statement, then I would be very grateful ! Or if someone knows who I can ask for this type of thing

I don't know what the "$(\infty,2)$" at the end of that message is doing here...

ah it wasn't there.... weird

@SaalHardali yeah this makes sense. This is sort of like the picture I was drawing when I was trying to understand the definition.

@MaximeRamzi Yeah, I agree it's clear what to do on objects / 1-categorically. I'm just hesitant to conclude that this yields a functor $\infty$-categorically. I believe it does, but I don't know how to show it.

@TimCampion Then I don't see what you mean by functorial; where does he claim that it is ? functorial in what ?

I mean, we start off with a functor $K \to X$, and end up with a functor $K \to X_s$, which Lurie calls $q'$

Yeah but he explains how to get it

Maybe i need to read about cocartesian fibrations more. I'm used to "cocartesian lift" just being used for single arrows

Yes !

That's what it is, just a single arrow in the functor category

The single arrow is the arrow $\Delta^1\to Fun(K,S)$ determined by the functor $Q: \Delta^1\times K\to S$ he describes :)

The point is that you have, in $S$, a cocone, which corresponds to this arrow $\Delta^1\to Fun(K,S)$. Now $Fun(K,X)\to Fun(K,S)$ is also a cocartesian fibration, which can be thought of as saying "cocartesian lifts are functorial" , this implies (because the domain of the arrow is $p\circ q$) that there is a lift to $Fun(K,X)$ whose domain is $q$ and whose codomain lives in $X_s$ (because the codomain of the original arrow was $s$)

i see. I did not catch that he was implicitly currying / uncurrying

and i didnt think of it myself either!