@bbb The description of the fibers follows from the fact that if $X$ is a simplicial set mapping into $\Delta^{1}$, then the $n$-simplices in the fibre $X \times _{\Delta^{1}} \{ i \}$ are in one-to-one correspondence with maps of simplicial sets $\Delta^{n} \rightarrow X$ over $\Delta^{1}$, where we equip $\Delta^{n}$ with the structure map given as the composite $X \rightarrow \Delta^{0} \rightarrow \Delta^{1}$, the second map being the inclusion of the $i$-th vertex.

When $X$ is Lurie's fundamental correspondence, you can read off the set of such maps from $\Delta^{n}$ straight from the universal property.

To answer your second question, in reasonable (ie. presentable, of which simplicial sets over $\Delta^{1}$ is an example) category $D$, a presheaf $F: D^{op} \rightarrow Set$ is representable (ie. there exists a - necessarily unique - object $X$ such that $Hom(-, X) \simeq F(-)$) if and only if it takes colimits in $D$ to limits in sets. In 7.3.6.4, the relevant presheaf is $F(A) = Hom(\overline{A}, C)$, and one can check directly it has this property.

When, in the course of human events, I'm led to delving into an involved description of a simplicial set in HTT, I invariably spend an inordinate amount of time reworking the definition until it's entirely written in terms of the 3 basic functors $f_! \dashv f^\ast \dashv f_\ast : sSet / X \to sSet / Y$ for $f: X \to Y$ -- the least obvious one to spot, of course, is always $f_\ast$. This rarely buys me anything concrete beyond functoriality, but it's reassuring nonetheless.

But I've finally found one which stumps me. In HTT 5.4.4.2, I can't seem to write down the simplicial set $T(K)$ (in the latest revision) in these terms.

(Here $f^\ast$ is pullback, $f_!$ is postcomposition, and $f_\ast$ has to do with the local cartesian closure of the 1-category $sSet$.)

Am I the only one hung up on this pointless repeated exercise? And if not, am I missing something in this particular instance?

This particular instance is in turn motivated by this question, which is probably equally pointless, being an exercise in "regular cardinal - golf". I like to justify myself by saying "but if $\kappa = \aleph_0$, then playing golf may be meaningful!" :)

@PiotrPstrągowski Thank you! Worked out the fiber computation you suggested and it makes complete sense to me now. Though I'll have to to admit I'm a little bummed that this didn't feel "obvious" to me, i.e. I probably wouldn't have came up with this proof on my own. I probably just need more practice.

Re: the second part - thanks for the explanation - I know this is basic ordinary category theory that I should know but do you know somewhere where I can find this representability criterion for presheaves on (ordinary) presentable categories written down?

@Bbb Not quite sure what would be the best source if you're looking for something elementary, but this fact is a form of the adjoint functor theorem, see ncatlab.org/nlab/show/adjoint+functor+theorem and references therein.

@PiotrPstrągowski Something that it is obvious when you write it down, but for some reason is almost never mentioned is that functor C^{op}→Set (or Space) is representable iff it is a right adjoint

(I know you know it, I'm just writing it for whoever could be listening)

@JonathanBeardsley after thinking about it a little, and kinda related to what Denis just written, the statement Mon(C_{\le n}) = Mon(C)_{\le n} only require C to have all small colimits. this is due to the slightly more general fact that E(D(C))=(E(x)D)(C) where E(C) is E-objects in C and E is presentable, while C is only in Cat^L.

Not that its that important, but in case you ever need it in this generality :-)

@PiotrPstrągowski ahh of course, had a brain fart, got it now, thanks

@IanColey i want to say that end(f) = end(x) \times end(y) characterizes it as a zero morphism but i can't seem to come up with an argument

Sorry, that statement isn't well-typed. I mean end(f) = end(id_x) \times end(id_y)

@IanColey If you allow yourself to talk about morphisms in $Ar(C)$ which are pushout squares in $C$, then there's a (pushout square, square whose top morphism is an equivalence) factorization system on $Ar(C)$. If you allow yourself to talk about objects of $Ar(C)$ whose codomain (as a morphism in $C$) is $0$, then $f \in Ar(C)$ is a zero morphism in $C$ iff it admits a map in the right half of the factorization system from an object whose codomain is 0. Not sure if that helps.

@IanColey ah the argument is simple: if end(f) = end(id_x) times end(id_y), then $f = f \circ id_x = 0_y \circ f = 0_{x \to y}$