Proof

Left to the reader.

Proof 1

It suffices to prove the result in the case of a diagram in the category of sets, in which case the proof is left to the reader.

Proof 2

Let us suppose that the ambiant category $\mathbf{C}$ has pullbacks. Then the category $\mathbf{C}^{I}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of pullback squares by lemma . Hence the class $\mathcal{R}$ is closed under composition and has the left cancellation property by the proposition here. The diagram can be represented by two maps $u:A\to B$ and $v:B\to C$ in the category $\mathbf{C}^{I}$. We have $v\in \mathcal{R}$ by assumption. It follows that $u\in \mathcal{R}$ iff $vu \in \mathcal{R}$, since the the class $\mathcal{R}$ is closed under composition and has the left cancellation property.

Proof

If $\mathbf{C}$ denotes the ambiant category, then the cube (viewed from high above) is a commutative square in the arrow category $\mathbf{C}^{I}$,

The faces $p:A\to B$ and $u:B\to D$ are cartesian by hypothesis. Hence also their composite $u p=v q$ by Lemma . It follows that the back face $q:A\to C$ is cartesian by the same lemma, since the right hand face $p:C\to D$ is cartesian by hypothesis.

Proof1

The category $\mathbf{E}=\mathbf{Set}^{I}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of cartesian squares by lemma . But the right class $\mathcal{R}$ of a factorisation system in a category $\mathbf{E}$ spans a full reflective subcategory of $\mathbf{E}^I$ by the proposition here. This proves that the full subcategory of $[I\times I, \mathbf{Set}]$ spanned by the cartesian squares is reflective, hence closed under limits.

Proof1

The category $I\times I$ is a projective cone $0\star C$ based on the category $C$ with three objects $\{1,2,3\}$ and two arrows $1\rightarrow 3\leftarrow 2$,

A square $S:I\times I\to \mathbf{Set}$ is cartesian iff the projective cone that it defines $0\star C\to \mathbf{Set}$ is exact. It is a general fact, valid for any category $C$, that the full subcategory of $[0\star C, \mathbf{Set}]$ spanned by the exact projective cones is reflective.

Proof

By diagram chasing.

Proof

The cube viewed from high above is a commutative square in the arrow category ${Set}^{I}$,

The faces $p:A\to B$ and $u:B\to D$ are epicartesian by hypothesis, hence also their composite $u p=v q$ by Lemma . It follows that the back face $q:A\to C$ is epicartesian by the same lemma, since the right hand face $p:C\to D$ is cartesian by hypothesis.

Proof

A retract $S'$ of a set $S$ is the set of fixed points of an idempotent map $e:S\to S$. Similarly, a retract $X'$ of a square $X$,

is the square of fixed points of an idempotent morphism $e:X\to X$. Let us show that $X'$ is epicartesian when $X$ is epicartesian. Let $x\in X'_{01}$ and $y\in X'_{10}$ be a pair of elements such that $q_1(x)=p_1(y)$. Then there exists an element $z\in X_{00}$ such that $p_0(z)=x$ and $q_0(z)=y$, since the square $X$ is epicartesian by assumption. But then $e(z)\in X'_{00}$, and we have $p_0(e(z))= e(p_0(z))=e(x)=x$, $q_0(e(z))= e(q_0(z))=e(y)=y$. This shows that the square $X'$ is epicartesian

Proof

Let $\alpha \gt 0$ be a limit ordinal, and let

be a continuous opchain with a base in the sub-category of surjections. This last condition means that the “restriction map” $D(i,j):D(i)\to D(j)$ is surjective for every $i\ge j\lt \alpha$. Let us show that the restriction map $D(\alpha,0):D(\alpha)\to D(0)$ is surjective. Let $P=D/[\alpha]$ be the poset of elements of the presheaf $D:[\alpha]^o \to \mathbf{Set}$. More precisely, an element of $P$ is a pair $(x,i)$ where $i\in [\alpha]$ and $x\in D(i)$. By definition, we have $(x,i)\leq (y,j)$ if $i\ge j$ and $D(i,j)(x)=y$. It follows from the continuity of $D$ that the poset $P$ is co-inductive (ie every chain in $P$ has an infumum). Moreover the minimal elements of $P$ are of the form $(z,\alpha)$ for $z\in D(\alpha)$ since the map $D(i)\to D(j)$ is surjective for every $\alpha \gt i\le j$.
It follows by Zorn lemma that for every $x\in D(0)$ there exists an element $z\in D(\alpha)$ such that $D(\alpha,0)(z)=x$.