Natural equivalence in higher category Let $C,D$ be two quasi category, we then have the functor category Fun($C,D$), a natural equivalence is just an equivalence in this functor category.
Given a natural equivalence $f: C\times\Delta^1\to D$, then the evaluation $f(id_c,id_{[1]})$ at every $c\in C$ is always an equivalence in $D$.
My question is, is the converse true? How can we prove it? In fact, I have no idea how to construct the inverse $g$ of $f$.
References or any idea about this is highly welcome. 
 A: This is true but nontrivial. Here is a sketch of the proof: construct the model structure on marked simplicial sets and show that it is Cartesian, with fibrant objects the naturally marked quasicategories. The Cartesian property implies in particular that the exponential of a fibrant object by a cofibrant object remains fibrant. That is, if $Q$ is a naturally marked quasicategory (viewed as a marked simplicial set) and $A$ is any marked simplicial set (everything is cofibrant) then the exponential $Q^A$ is a naturally marked quasicategory. 
However, the marking on $Q^A$ makes an edge $h: A\times \Delta^1 \to Q$ marked if and only if it sends each pair $(f,\mathrm{id}_{[1]})$, where $\mathrm{id}_{[1]}$ is the nontrivial edge of $\Delta^1$ and $f$ is marked in $A$, to a marked edge (that is, an equivalence) in $Q$. It is not too hard to reduce this using composition in $Q$ to the claim that $h(\mathrm{id}_a,\mathrm{id}_{[1]})$ is marked for every degenerate 1-simplex $\mathrm{id}_a$ in $A$, which is exactly the claim that $h$ is a pointwise equivalence. 
So, $Q^A$ has as marked edges the pointwise equivalences, by the construction of the exponential, and must also be a naturally marked quasicategory, so that the marked edges are the legitimate equivalences, and the two classes coincide as desired. I do not know an argument avoiding marked simplicial sets, although Joyal probably gives one. For more details of this argument you can look in Riehl-Verity, The 2-category theory of quasicategories, section 2.3. They leave the construction of the model structure itself to Lurie. 
