Factorization systems

Proof. These properties all follow from the definition of orthogonality in terms of hom animae. The proofs of (1) and (2) are left to the reader. For (3), consider a pullback square in \(C\) where \(r \in S^\perp \):

For any \(l \colon A \to B\) in \(S\), we must show \(l \perp r'\). This means showing that the left square in the diagram below is a pullback square of animae:

The right square is a pullback because \(\Hom _C(A,-)\) preserves pullbacks, so by the pasting lemma for pullback squares it remains to show that the outer rectangle is a pullback. But this holds as we may rewrite it as the pasting of the square obtained from the original pullback square by applying \(\Hom _C(B,-)\) and the pullback square exhibiting the orthogonality relation \(l \perp r\). The argument for (4) is dual.

For (5), consider morphisms \(f\colon X \to Y\) and \(g\colon Y \to Z\) in \(C\), and assume that \(g \circ f\) and \(g\) lie in \(S^{\perp }\). To show that \(f \in S^{\perp }\), let \(l\colon A \to B\) be a morphism in \(S\), and consider the following commutative diagram:

By assumption, the right and outer squares are pullback squares, hence by the pasting law so is the left square, as was to be shown. The argument for (6) is dual.

For (7), let \(f \colon I \to \Ar (C)\) be a diagram where each morphism \(f(i)\colon X(i) \to Y(i)\) lies in \(S^\perp \). Let \(r \colon \lim _i X(i) \to \lim _i Y(i)\) be the limit morphism. For any \(l \colon A \to B\) in \(S\), we have a chain of equivalences: \begin {align*} \Hom _C(B, \lim _i X(i)) &\simeq \lim _i \Hom _C(B, X(i)) \\ \Hom _C(A, \lim _i X(i)) &\simeq \lim _i \Hom _C(A, X(i)) \\ \Hom _C(B, \lim _i Y(i)) &\simeq \lim _i \Hom _C(B, Y(i)) \\ \Hom _C(A, \lim _i Y(i)) &\simeq \lim _i \Hom _C(A, Y(i)). \end {align*}

The square defining the orthogonality \(l \perp r\) is the limit of the squares defining \(l \perp f(i)\) for each \(i \in I\). Since limits of pullback squares are pullback squares, the claim follows. The proof for (8) is dual. □

Proof. By symmetry, it suffices to prove \(\Rr = \Ll ^\perp \). The inclusion \(\Rr \subseteq \Ll ^\perp \) holds by definition. For the converse, let \(r \colon X \to Y\) be a morphism in \(\Ll ^\perp \). By the factorization property, we can write \(r \simeq r' \circ l\) for some \(l \colon X \to Z\) in \(\Ll \) and \(r' \colon Z \to Y\) in \(\Rr \).

Since \(l \in \Ll \) and \(r \in \Ll ^\perp \), we have \(l \perp r\). This means there is a contractible anima of lifts in the commutative square

Let \(d \colon Z \to X\) be such a lift. The commutativity of the top triangle gives \(d \circ l \simeq \mathrm {id}_X\). To see that also \(l \circ d \simeq \mathrm {id}_Z\), consider the following lifting problem for \(l \perp r'\):

It is clear that \(\id _Z\colon Z \to Z\) is a filler. But also the morphism \(l \circ d \colon Z \to Z\) is a filler, since \(r' \circ (l \circ d) \simeq (r' \circ l) \circ d \simeq r \circ d \simeq r'\) and \((l \circ d) \circ l \simeq l \circ (d \circ l) \simeq l\). By uniqueness of fillers, we must have \(l \circ d \simeq \mathrm {id}_Z\).

We conclude that \(l\) is an isomorphism, hence in \(\Rr \). Since \(r \simeq r' \circ l\) and \(\Rr \) is closed under composition, we conclude that also \(r \in \Rr \). □

Proof. Assume first that \((\Ll ,\Rr )\) is a factorization system. The existence of factorizations implies that \(- \circ -\) is essentially surjective, so it remains to show it is also fully faithful. To this end, consider morphisms \(l\colon X \to Z\) and \(l'\colon X' \to Z'\) in \(\Ll \), and morphisms \(r\colon Z \to Y\) and \(r'\colon Z' \to Y'\) in \(\Rr \), and define \(u := r \circ l\) and \(u' := r' \circ l'\). We must show that the map \[ \Hom _{\Fact _{\Ll ,\Rr }(C)}((l,r),(l',r')) \to \Hom _{\Ar (C)}(u,u') \] induced by composition in \(C\) is an equivalence of animae. Equivalently, we may show that the fiber over every morphism \(u \to u'\) in \(\Ar (C)\) is contractible. Such a morphism consists of maps \(a\colon X \to X'\) and \(b\colon Y \to Y'\) in \(C\), together with a homotopy filling the outer rectangle in the following diagram:

Using the commutative-square and Segal axioms, we then compute that the fiber under consideration agrees with the fiber over \((l'a\colon X \to Z', br\colon Z \to Y')\) of the map \[ (l^*, r'_*)\colon \Hom _C(Z, Z') \to \Hom _C(X,Z') \times _{\Hom _C(X,Y')} \Hom _C(Z, Y'). \] But this map is an equivalence, since \(l \perp r'\).

The proof of the converse is similar: if the composition functor is an equivalence, then essential surjectivity provides the desired factorizations of morphisms in \(C\), while full faithfulness applied to the objects \((l\colon A \to B, \id _{B}\colon B \to B)\) and \((\id _X\colon X \to X, r\colon X \to Y)\) of \(\Fact _{\Ll ,\Rr }(C)\) provides the orthogonality condition. □

Proof. Given \(r\) and \(g\), factor the composite \(gr\) as \(r'f\) with \(f\in \Ll \) and \(r'\in \Rr \). This provides a lift of \(g\) of the displayed form. We show that every such lift is \(t\)-cocartesian.

Consider a third object \((r''\colon x''\to y'')\) in \(\Ar _{\Rr }(C)\). We must show that the square

is a pullback square. Fix a morphism \(g'\colon y'\to y''\) and a morphism \((h,g'g)\colon r\to r''\) over \(g'g\). The fiber in question is the anima of maps \(k\colon x'\to x''\) fitting into the diagram

Indeed, the two required triangles say precisely that \(kf\simeq h\) and \(r''k\simeq g'r'\). This filler anima is contractible because \(f\perp r''\). It follows that the displayed square of hom animae is a pullback, so the original morphism is \(t\)-cocartesian.

Conversely, every \(t\)-cocartesian lift of \(g\) with source \(r\) is isomorphic to one constructed by factoring \(gr\). Its top morphism is therefore the composite of a morphism in \(\Ll \) with an isomorphism, and hence lies in \(\Ll \). □

Proof. Any span \(X \xleftarrow {l} U \xrightarrow {r} Y\) can be factored as a backwards map followed by a forwards map: \[ X \xleftarrow {l} U \xrightarrow {\mathrm {id}_U} U \qquad \text {followed by} \qquad U \xleftarrow {\mathrm {id}_U} U \xrightarrow {r} Y. \] It remains to check orthogonality. Let \(l' \colon B \to A\) be a backwards map (given by \(l \in C_L\)) and \(r' \colon X \to Y\) be a forwards map (given by \(r \in C_R\)). We need to show that the square of hom animae

is a pullback. Using the formula for hom animae in a span category (see Lemma 13.1.13), this square is equivalent to

The square involving only \(C_L\) is degenerate, hence a pullback square, and similarly for the square involving \(C_R\). We conclude that this square is a pullback square, finishing the proof. □

Proof. Both classes define wide subcategories of \(E\): they contain all isomorphisms, and they are closed under composition because \(\Ll \) and \(\Rr \) are, and because composites of \(p\)-cocartesian morphisms are again \(p\)-cocartesian.

(1) For orthogonality, consider \(\tilde {l}\colon \widetilde {A} \to \widetilde {B}\) in \(\widetilde {\Ll }\) and \(\tilde {r}\colon \widetilde {X} \to \widetilde {Y}\) in \(\widetilde {\Rr }\). Let \(l\colon A \to B\) and \(r\colon X \to Y\) be their images in \(C\). We need to show that the top square in the following commutative cube is a pullback square:

Since \(\widetilde {l}\) is \(p\)-cocartesian, the left and right squares are pullback squares, and since \(l \perp r\) the bottom square is a pullback square. The claim thus follows from the pasting law of pullback squares.

(2) For factorization, consider a morphism \(\widetilde {f}\colon \widetilde {X} \to \widetilde {Z}\). We may factor its image \(X \to Z\) in \(C\) as a composite of a morphism \(l\colon X \to Y\) in \(\Ll \) and a morphism \(r\colon Y \to Z\) in \(\Rr \). We may then take \(\widetilde {l}\colon \widetilde {X} \to \widetilde {Y}\) to be a \(p\)-cocartesian lift of \(l\), which exists by assumption. The cocartesian property produces a unique morphism \(\widetilde {r}\colon \widetilde {Y} \to \widetilde {Z}\) over \(r\) through which \(\widetilde {f}\) factors, as desired. □

Proof. This is an instance of Lemma 17.1.12 applied to the functor \(p_{\Oo }\colon \Oo ^{\otimes } \to \Span (\Fin )\), where we equip the target with the factorization system from Proposition 17.1.10. □