Morphisms in category of left fibrations

Question

I am trying to better understand the straightening-unstraightening equivalence of Lurie in the $\infty$-categorical setting. In the case that I am interested in, this equivalence states that $$ \mathrm{LFib}(\mathcal{C}) \simeq \mathrm{Fun}(\mathcal{C}, \mathrm{Spaces}), $$ where $\mathcal{C}$ is an $\infty$-category and $\mathrm{LFib}(\mathcal{C})$ is some $\infty$-subcategory of $(\mathrm{Cat}_\infty)_{/\mathcal{C}}$ spanned by the left fibrations with codomain $\mathcal{C}$. What I don't understand and can't quite find in the literature is whether the morphisms in $\mathrm{LFib}(\mathcal{C})$, which I take to be functors $\mathcal{D} \to \mathcal{E}$ sitting over $\mathcal{C}$, are themselves left fibrations. I.e., if I have a natural transformation $\eta : D \to E$ of functors $D, E : \mathcal{C} \to \mathrm{Spaces}$, does it straighten to a left fibration $\mathcal{D} \to \mathcal{E}$ between the straightenings of $D$ and $E$?

Answer 1

The answer is yes, but $\mathrm{LFib}(\mathcal C)$ is also the full subcategory of $(\mathrm{Cat}_\infty)_{/\mathcal C}$, it just so happens that you can prove that any morphism between such is a left fibration (this does not remain true in the case of cocartesian fibrations, though).

Here is a proof: Let $f:\mathcal{D\to E}$ be a morphism of left fibrations over $\mathcal C$, which I'll denote by $p,q$ respectively. By the dual of HTT.2.4.1.3.(3), if an edge $\alpha$ in $\mathcal D$ is such that $f(\alpha)$ is $q$-coCartesian, then $\alpha$ is $f$-coCartesian if and only if it is $p$-coCartesian.

But we are in left fibrations, so all edges are $p$-coCartesian. In particular, $\alpha$ is $f$-coCartesian if and only if $f(\alpha)$ is $q$-coCartesian, the latter being a void condition: every edge is $f$-coCartesian.

So we have seen that any edge is $f$-coCartesian, and therefore we just need a sufficient supply of (arbitrary) edges along $f$: let $\alpha: e_0\to e_1$ be an edge in $\mathcal E$, with a lift $d_0$ of $e_0$ along $f$, i.e. $f(d_0) = e_0$. We then push things down: $q(\alpha): q(e_0) \to q(e_1)$ is an edge in $\mathcal C$, with source $q(e_0) = p(d_0)$, so it has a lift $\tilde \alpha: d_0\to d_1$. Now $f(\tilde\alpha): f(d_0)\to f(d_1)$ is some map lifting $q(\alpha)$, and it is $q$-coCartesian, just like $\alpha$ (and any map in $\mathcal E$), so there is an equivalence $e_1\simeq f(d_1)$ with appropriate $2$-simplices showing that have found a lift of $\alpha$.