ScalingStacks

Proof. By symmetry, it is enough to prove (1). Observe that the prism $\Delta^{1}\times\Delta^{2}$ is a left cone on the simplicial set obtained by removing the initial vertex. Formally,

$$\Delta^{1}\times\Delta^{2}\simeq\left(\Delta^{2}\times\Delta^{\left\{1\right\}}\sqcup_{\Delta^{\left\{1,2\right\}}\times\Delta^{\left\{1\right\}}}\Delta^{\left\{1,2\right\}}\times\Delta^{1}\right)^{\triangleleft}.$$

We can therefore interpret the rectangle as a diagram in $\mathcal{C}_{A/}$ (and hence ignore $A$). Since the projection $\mathcal{C}_{A/}\to\mathcal{C}$ preserves and reflects limits (dual of T.1.2.13.8), the square $q_{r}$ is a pullback square in $\mathcal{C}_{A/}$. The universal property of the pullback implies that we have a homotopy Cartesian square

which in turn induces a homotopy equivalence of homotopy fibers of the vertical maps. Considering the given map $B\to Y$ as a point in $\operatorname{Map}_{\mathcal{C}_{A/}}\left(B,Y\right)$ and considering the induced equivalence on the homotopy fibers of the vertical maps, we obtain by T.5.5.5.12 an equivalence

$$\operatorname{Map}_{\mathcal{C}_{A//Y}}\left(\overline{B},\overline{X}\right)\overset{\sim}{\longrightarrow}\operatorname{Map}_{\mathcal{C}_{A//W}}\left(\overline{\overline{B}},\overline{\overline{Z}}\right),$$

where $\overline{B}$ and $\overline{X}$ are $A\to B\to Y$ and $A\to X\to Y$ viewed as objects of $\mathcal{C}_{A//Y}$ and $\overline{\overline{B}}$ and $\overline{\overline{Z}}$ are $A\to B\to W$ and $A\to Z\to W$ viewed as objects of $\mathcal{C}_{A//W}$. By the definition of the space of lifts, this is precisely the equivalence $L\left(q_{l}\right)\simeq L\left(q\right)$. ∎

Proof. Let $\mathcal{M}\to\Delta^{1}$ be the Cartesian-coCartesian fibration associated with the adjunction $F\dashv G$. Since $\mathcal{C}$ and $\mathcal{D}$ are full subcategories of $\mathcal{M}$ we can think of the square $q$ as taking values in $\mathcal{M}$ and it does not change the space of lifts. Consider the diagram in $\mathcal{M}$ given by

where in the left square $q_{l}$ the horizontal arrows are coCartesian and the rest of the data is given by the lifting property of coCartesian edges. Since the inclusion of the spine $\Lambda_{1}^{2}\hookrightarrow\Delta^{2}$ is inner anodyne, so is $\Delta^{1}\times\Lambda_{1}^{2}\hookrightarrow\Delta^{1}\times\Delta^{2}$ (by T.2.3.2.4) and since $\mathcal{M}\to\Delta^{1}$ is an inner fibration, the diagram can be extended to $\Delta^{1}\times\Delta^{2}\to\mathcal{M}$ and we can denote the outer square by $r\colon\Delta^{1}\times\Delta^{1}\to\mathcal{M}$. We now claim that $q_{l}$ is a pushout square in $\mathcal{M}$. For every $Z\in\mathcal{M}$, consider the induced diagram

If $Z\in\mathcal{M}_{0}\simeq\mathcal{C}$, then the spaces on both left corners are empty and if $Z\in\mathcal{M}_{1}\simeq\mathcal{D}$, then both horizontal arrows are equivalences. Either way, this is a pullback square and hence $q_{l}$ is a pushout square. By 4.1.3 we get $L\left(q\right)\simeq L\left(r\right)$.

We can now factor the outer square $r\colon\Delta^{1}\times\Delta^{1}\to\mathcal{M}$ as

where the left square is $p$ and in the right square $q_{r}$ the horizontal arrows are Cartesian and the square is determined by the lifting property of Cartesian edges. Repeating the argument in the dual form we get that $q_{r}$ is a pullback square and using 4.1.3 again we get $L\left(p\right)\simeq L\left(r\right)$ and therefore $L\left(p\right)\simeq L\left(q\right)$. ∎

Proof. For ease of notation, set $\mathcal{D}=\mathcal{C}_{A/}$. Recall that

$$L\left(q\right)=\operatorname{Map}_{\mathcal{D}_{/\overline{Y}}}\left(\overline{B},\overline{X}\right)$$

and therefore

$$L\left(q\right)\times L\left(q\right)=\operatorname{Map}_{\mathcal{D}_{/\overline{Y}}}\left(\overline{B},\overline{X}\right)\times\operatorname{Map}_{\mathcal{D}_{/\overline{Y}}}\left(\overline{B},\overline{X}\right)\simeq\operatorname{Map}_{\mathcal{D}_{/\overline{Y}}}\left(\overline{B},\overline{X}\times\overline{X}\right).$$

Products in the over-category are fibered products and products in the under-category are just ordinary products (dual of T.1.2.13.8). Hence, $\overline{X}\times\overline{X}$ is the diagram $A\to X\times_{Y}X\to Y$, which we denote by $\overline{X\times_{Y}X}$. Thus, a point $s=\left(s_{0},s_{1}\right)\in L\left(q\right)\times L\left(q\right)$ corresponds to a lift in the diagram

in the category $\mathcal{D}$. Furthermore, the diagonal map $\delta_{L\left(q\right)}\colon L\left(q\right)\to L\left(q\right)\times L\left(q\right)$ is induced from the diagonal map $\delta_{\overline{X}}\colon\overline{X}\to\overline{X}\times\overline{X}$. Namely, $\delta_{L\left(q\right)}=\left(\delta_{\overline{X}}\right)_{*}$. Our goal is therefore to compute the homotopy fiber of $\left(\delta_{\overline{X}}\right)_{*}$ over a given point

$$s=\left(s_{0},s_{1}\right)\simeq\operatorname{Map}_{\mathcal{D}_{/\overline{Y}}}\left(\overline{B},\overline{X}\times\overline{X}\right).$$

The projection $\mathcal{D}_{/\overline{Y}}\to\mathcal{D}$ induces an equivalence

$$\left(\mathcal{D}_{/\overline{Y}}\right)_{/\overline{X\times_{Y}X}}\simeq\mathcal{D}_{/\overline{X\times_{Y}X}}.$$

It follows that the fiber is the space of lifts in the diagram

in $\mathcal{D}$. By (the dual of) T.5.5.5.12, this space of lifts is homotopy equivalent to the mapping space $\operatorname{Map}_{\mathcal{D}_{/\overline{X\times_{Y}X}}}\left(\overline{B},\overline{X}\right)$. Recalling that $\mathcal{D}=\mathcal{C}_{A/}$, we see that this is none other than the space of lifts for $p$. ∎

Proof. As a right adjoint, $G$ is left exact and therefore preserves $d$-truncated morphisms by T.5.5.6.16. Since $G$ preserves $n$-truncated morphisms and the space of lifts in the square

is homotopy equivalent to the space of lifts in the adjoint square

given by 4.1.4, we see that if $f$ is left orthogonal to all $n$-truncated morphisms then so is $F\left(f\right)$. ∎

Proof. Since $\mathcal{C}$ has all pullbacks, every commutative square $q\colon\Delta^{1}\times\Delta^{1}\to\mathcal{C}$ of the form

can be factored as

By 4.1.3, the space of lifts for the original square $q$ is equivalent to the space of lifts in the left square of the above rectangle. Moreover, $n$-truncated morphisms are closed under base change and so to check that $f$ is $n$-connected, we can equivalently restrict ourselves to checking the left orthogonality condition only for squares $q$ in which the map $B\to Y$ is the identity on $B$. Writing $\overline{A}$, $\overline{X}$ and $\overline{B}$ for $A\to B$, $X\to B$ and $\operatorname{Id}\colon B\to B$ as objects of $\mathcal{C}_{/B}$, respectively, we see that by the dual of T.5.5.5.12 the space of lifts fits into a fiber sequence

$$L\left(q\right)=\operatorname{Map}_{\mathcal{C}_{A//B}}\left(\overline{B},\overline{X}\right)\to\operatorname{Map}_{\mathcal{C}_{/B}}\left(\overline{B},\overline{X}\right)\xrightarrow{f^{*}}\operatorname{Map}_{\mathcal{C}_{/B}}\left(\overline{A},\overline{X}\right).$$

Hence, $f$ is $n$-connected if and only if $f^{*}$ is an equivalence for every $n$-truncated morphism $X\to B$. By T.5.5.6.10, a morphism $X\to B$ is $n$-truncated if and only if $\overline{X}$ is an $n$-truncated object of $\mathcal{C}_{/B}$. Hence, we need the above map to be an equivalence for every $n$-truncated object $\overline{X}\in\mathcal{C}_{/B}$. This precisely means that the map $\overline{A}\to\overline{B}$ exhibits $\overline{B}$, the terminal object of $\mathcal{C}_{/B}$, as the $n$-truncation of $\overline{A}$. ∎

Proof. We prove this by induction on $d$. For $d=n$, the claim follows from the definition of an $n$-connected morphism and the fact that a space is $\left(-2\right)$-connected if and only if it is contractible. We now assume that this is true for $d-1$, and prove it for $d$. Denote the space of lifts by $L\left(q\right)$. By T.5.5.6.15, it suffices to show that the diagonal map $\delta\colon L\left(q\right)\to L\left(q\right)\times L\left(q\right)$ is $\left(d-n-3\right)$-truncated. By 4.1.5, the homotopy fiber over a point $\left(s_{0},s_{1}\right)\in L\left(q\right)\times L\left(q\right)$ is equivalent to the space of lifts in the square

where the bottom map is $\left(s_{0},s_{1}\right)$. By T.5.5.6.15, since $X\to Y$ is $d$-truncated, $X\to X\times_{Y}X$ is $\left(d-1\right)$-truncated and, therefore, by induction, the space of lifts is $\left(\left(d-1\right)-n-2\right)$-truncated and we are done. ∎

Proof. By the Yoneda lemma it is enough to show that for every $n$-truncated object $Z$ in $\mathcal{C}$ the induced map

$$f_{*}\colon\operatorname{Map}\left(B,Z\right)\to\operatorname{Map}\left(A,Z\right)$$

is an equivalence. For this, it is enough to show that for every $g\colon A\to Z$, the fiber of $f_{*}$ over $g$ is contractible. By T.5.5.5.12, the fiber is equivalent to the space of lifts for the square

which is contractible by definition as $f\colon A\to B$ was assumed to be $n$-connected. ∎

Proof. For $\mathcal{C}=\mathcal{S}$, this follows from inspecting the induced map between the long exact sequences of homotopy groups associated with the vertical maps. For $\mathcal{C}=\mathcal{S}^{K}$, this follows from the claim for $\mathcal{S}$, since both truncation and pullbacks are computed level-wise. A general $\infty$-topos is a left exact localization of $\mathcal{S}^{K}$ for some $K$, and left exact colimit-preserving functors between presentable $\infty$-categories commute with truncation by T.5.5.6.28 and with pullbacks by assumption. Finally, by T.6.4.1.5 every $m$-topos is the full subcategory on $\left(m-1\right)$-truncated objects in an $\infty$-topos and this full subcategory is closed under limits. ∎

Proof. To show that $f\colon A\to B$ is $n$-connected, we need to show that the space of lifts for every square

in which the right vertical arrow is $n$-truncated, is contractible. Applying 4.3.3 and 4.1.3, we see that this space is equivalent to the space of lifts in the square

which, by 4.1.4, is equivalent to the space of lifts in the adjoint square

which is contractible since the left vertical arrow is an equivalence. ∎

Proof. We first prove the case of $m=\infty$. For $n=-2$, there is nothing to prove, and so we assume that $n\geq-1$. Since $f\circ s=\operatorname{Id}_{B}$ we get $\tau_{\leq n}^{\mathcal{C}}\left(f\right)\circ\tau_{\leq n}^{\mathcal{C}}\left(s\right)=\operatorname{Id}_{B}$ and since $\tau_{\leq n}\left(f\right)$ is an equivalence, then so is $\tau_{\leq n}\left(s\right)$ and hence $s$ is $\left(n-\frac{1}{2}\right)$-connected. By 4.3.4, $s$ is $\left(n-1\right)$-connected and hence, by T.6.5.1.20, the map $f$ is $n$-connected (note that $n$-connective means $\left(n-1\right)$-connected).

For a general $m$, by T.6.4.1.5 there exists an $\infty$-topos $\mathcal{D}$ and an equivalence $\mathcal{C}\simeq\tau_{\leq m-1}\mathcal{D}$, and so we may identify $\mathcal{C}$ with the full subcategory of $\left(m-1\right)$-truncated objects of $\mathcal{D}$. If $f\colon A\to B$ is $\left(n-\frac{1}{2}\right)$-connected in $\mathcal{C}$, then it is also $\left(n-\frac{1}{2}\right)$-connected in $\mathcal{D}$, since the restriction of $\tau_{\leq n}^{\mathcal{D}}$ to $\mathcal{C}$ is equivalent to $\tau_{\leq n}^{\mathcal{C}}$. It follows from the case of $m=\infty$ that $f$ is $n$-connected in $\mathcal{D}$. Since $f=\tau_{\leq m-1}^{\mathcal{D}}f$ and $\tau_{\leq m-1}^{\mathcal{D}}$ is a left adjoint functor, by 4.2.5 the map $f$ is also $n$-connected as a map in $\mathcal{C}$. ∎

Proof. Given a morphism $f\colon A\to B$ in $\mathcal{D}$, using the canonical simplicial resolution provided by the proof of A.4.7.3.13, we can express it as a colimit of the simplicial diagram of morphisms:

$$\operatorname*{colim}\limits_{\Delta^{op}}\left(T^{n+1}\left(A\right)\to T^{n+1}\left(B\right)\right),$$

which one can write as

$$\operatorname*{colim}\limits_{\Delta^{op}}\left(FT^{n}U\left(A\right)\to FT^{n}U\left(B\right)\right).$$

If $U\left(f\right)$ is $n$-connected as in the statement, then since $T$ preserves $n$-connected morphisms by assumption and $F$ preserves $n$-connected morphisms by being left adjoint, it follows that all the maps in the diagram are $n$-connected. By T.5.2.8.6(7), the map $f$ is also $n$-connected. ∎

Proof. Since $\mathcal{C}$ is presentable and the tensor product commutes with colimits separately in each variable, for each object $X\in\mathcal{C}$ the functor $Y\mapsto X\otimes Y$ is a left adjoint and therefore preserves $n$-connected morphisms by 4.1.4. Hence, given two $n$-connected morphisms $f\colon A_{1}\to B_{1}$ and $g\colon A_{2}\to B_{2}$, the composition

$$A_{1}\otimes B_{1}\xrightarrow{A_{1}\otimes g}A_{1}\otimes B_{2}\xrightarrow{f\otimes B_{2}}A_{2}\otimes B_{2}$$

is $n$-connected as a composition of two $n$-connected morphisms. ∎

Proof. By A.4.7.3.11, the adjunction $F\dashv U$ is monadic. Hence, given a morphism $A\to B$ in $\mathcal{C}$, by 2.4.6 the morphism $T\left(A\right)\to T\left(B\right)$ can be expressed as

$$\coprod_{n\geq 0}\left(P\left(n\right)\otimes A^{\otimes n}\right)_{h\Sigma_{n}}\to\coprod_{n\geq 0}\left(P\left(n\right)\otimes B^{\otimes n}\right)_{h\Sigma_{n}}.$$

By 4.4.2, $n$-connected morphisms are closed under $\otimes$ and, by T.5.2.8.6, they are closed under colimits. Hence, we obtain that $T\left(A\right)\to T\left(B\right)$ is $n$-connected as well. ∎

Proof. Combine 4.4.4 and 4.4.1. ∎