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We first justify the relevant pointwise epimorphism in a functor category criterion. A componentwise surjective function gives an epimorphism, because equality of two composites can be cancelled at each component. Conversely, for a natural transformation $t:X\to Y$, form the pointwise pushout $P=Y{⨿}_{X}Y$. Restrictions of $Y$ preserve the identifications imposed by $t$, so these pointwise sets and restrictions form a categorical presheaf. The two inclusions $j_1,j_2:Y\to P$ satisfy $j_{1}t=j_{2}t$. If $t$ is epic, $j_1=j_2$. At an object $U$, an element outside the image of $t_U$ would retain two distinct copies in $P(U)$, contradicting this equality. Thus every $t_U$ is surjective.

Apply this criterion to $H_{\bullet }(f)$ at $B$. Surjectivity of $\mathcal{C}(B,A)\to \mathcal{C}(B,B)$ supplies $g:B\to A$ with $fg=1_B$. Conversely, if $fg=1_B$, then $H_{\bullet }(f)H_{\bullet }(g)=1_{H_B}$, making $H_{\bullet }(f)$ a split epimorphism and hence an epimorphism. Therefore

Work in the frame $A$, writing $a\to x$ for Heyting implication. A nucleus on a frame is an inflationary, idempotent map preserving finite meets. The open nucleus and closed nucleus areTheir meet in the frame of nuclei on a frame is pointwise. The distributivity of a Heyting algebra and the defining adjunction for Heyting implication giveIf a nucleus on a frame $l$ bounds both, then $l(a)=1$ and $a\le l(0)$. Consequently $1=l(a)\le l(l(0))=l(0)$, so $l$ is the top nucleus. Thus $o(a)$ and $c(a)$ are complements in $N(A)$.

For any nucleus on a frame $j$, put $t_a=o(a)\wedge c(j(a))$. At an arbitrary $x$,since $j(a)\wedge (a\to x)\le j(a)\wedge j(a\to x)=j(a\wedge (a\to x))\le j(x)$. Moreover $t_a(a)=j(a)$. ThereforeIndeed the join is at most $j$, and its value at each $a$ is at least $j(a)$. This proves generation by the open nuclei and closed nuclei, without assuming that joins of nuclei on a frame are computed pointwise.

For the given frame homomorphism $h$, restrict the displayed join to $a=h(b)$ and denote the resulting nucleus on a frame by $r(j)$. The same inequalities show $r(j)\le j$ and $r(j)(h(b))=j(h(b))$. If $k$ has the same values on $h(B)$, its restricted formula is identical, so $r(j)\le k$. Hence $r(j)$ is the least representative of the equivalence class. We will also use the consequence

The least representatives form a subframe. For a family $j_i=r(j_i)$, let $J={\bigvee }_i{j}_i$ in $N(A)$. Restricted monotonicity gives $j_i\le r(J)$ for every $i$, hence $J=r(J)$. For two least representatives, distribution in $N(A)$ expresses their meet as the join, over $b,d\in B$, ofHere $o(hb)\wedge o(hd)=o(h(b\vee d))$ and $c(u)\wedge c(v)=c(u\wedge v)$. Sinceeach summand is below $r(j\wedge k)$. The opposite inequality always holds, proving closure under binary meets. The top is $c(1)$ and is a least representative. Also $r(c(a))=c(a)$, because the summand at $b=0$ is $c(a)$; and $r(o(hd))=o(hd)$, because the summand at $b=d$ is $o(hd)$. It follows from the restricted normal form that $N_B(A)$ is exactly the subframe generated by $c(A)$ and $o(h(B))$.

The map $c:A\to N_B(A)$ preserves finite meets by the formula above. A nucleus on a frame $l$ bounds every $c(a_i)$ exactly when $a_i\le l(0)$ for every $i$, equivalently ${\bigvee }_i{a}_i\le l(0)$. The least such nucleus is $c({\bigvee }_i{a}_i)$, so $c$ preserves arbitrary joins too, including $0$. Evaluation at $0$ recovers $a$, so this frame homomorphism is injective.

For the pushout, we prove the universal property explicitly. Let $u:A\to C$ be a frame homomorphism for which every $u(hb)$ has a complement $d_b$ in $C$. Define, for $j\in N_B(A)$,To prove preservation of arbitrary joins, let $u_{\ast }$ be the right adjoint of $u$ and set $k_z(a)=u_{\ast }(z\vee u(a))$ for $z\in C$. This is a nucleus on a frame: inflationarity follows from the adjunction, finite-meet preservation follows from distribution and meet preservation by $u_{\ast }$, and $u{u}_{\ast }\le 1$ gives $k_z{k}_z\le k_z$; inflationarity gives the reverse inequality. Complementation and the adjunction now giveThe last equivalence uses leastness of $j$. Thus $F$ is a left adjoint, so preserves arbitrary joins. For binary meets, expand $F(j)\wedge F(k)$. A term indexed by $b,d$ isThis proves $F(j)\wedge F(k)\le F(j\wedge k)$; monotonicity proves the reverse inequality. The term at $b=0$ proves $F(1)=1$. Directly,For the second identity the term at $b=d$ gives $d_d$, while every term is disjoint from $u(hd)$, because $hd\wedge (hd\to hb)\le hb$.

Apply this construction first with $A=B$, $h=1_B$, target $N_B(A)$, $u=c_{A}h$ and complements $d_b=o_A(hb)$. It gives a frame homomorphism $v_0:N(B)\to N_B(A)$ satisfying $v_0{c}_B=c_{A}h$ and $v_0{o}_B=o_{A}h$. Given any compatible pair $u:A\to C$, $v:N(B)\to C$, take $d_b=v(o_B(b))$; compatibility supplies the required complements. The constructed $F$ satisfies $F{c}_A=u$ and $F{v}_0=v$, since the latter equality holds on the open and closed generators of $N(B)$. Uniqueness follows because $c(A)$ and $o(h(B))$ generate $N_B(A)$. Hence the square with edges $h,c_B,c_A,v_0$ is a pushout in $\mathbf{Frm}$.

The van Kampen theorem says: if $X=U\cup V$ with $U,V$ open and path connected, $U\cap V$ path connected, and a base point in the intersection, then ${\pi }_1(X)$ is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of every element of ${\pi }_1(U\cap V)$. Equivalently it is the pushout of the two inclusion-induced fundamental-group maps.

A coherent category is a regular category with finite unions of subobjects, including the empty union, stable under pullback. Thus it has finite limits, pullback-stable regular-epimorphism/monomorphism image factorizations, and pullback-stable finite joins in each subobject lattice.

Write $U=A\vee B\hookrightarrow X$ and $C=A\wedge B=A{\times }_{X}B$. Their inclusions give a commutative squareTo establish the pushout property, take arrows $a:A\to Y$, $b:B\to Y$ agreeing on $C$. Regard their graphs as subobjects of $U\times Y$ and form their union $R$. Internally, this relation says that a point of $U$ is related to its $a$-value when it lies in $A$, or to its $b$-value when it lies in $B$.

The relation is total because every point of $U$ belongs to at least one of $A,B$. It is single-valued: two witnesses in the same summand give the same value by functionality of that map; witnesses in different summands have their shared point in $C$, where the two maps agree. By the relation argument proved in Question 7's unheaded continuation, the projection $R\to U$ is both a regular epimorphism and monic, hence an isomorphism. It consequently gives a map $h:U\to Y$ restricting to $a,b$.

For uniqueness, the equalizer of two possible maps contains $A$ and $B$, hence contains their union $U$ and is all of $U$. Thus the square is a pushout, provingThis is the unions of subobjects are pushouts property; it does not assume arbitrary pushouts already exist in a coherent category.

Use homological grading $d_n:C_n\to C_{n-1}$, with zero terms below degree zero. In the projective model structure on nonnegative chain complexes, the classes are:

Thus a cofibrant complex has a projective module in each degree, and every object is fibrant. An projective acyclic fibration is equivalently a degreewise surjective quasi-isomorphism, including degree zero. To see the degree-zero assertion, lift a degree-zero homology class using the homology isomorphism, then lift the boundary discrepancy using surjectivity in degree one. Conversely a degreewise surjective quasi-isomorphism has the requisite positive-degree surjections.

Let $S^n(R)$ be the sphere chain complex with $R$ in degree $n$ and zero elsewhere. Let $D^n(R)$, for $n\ge 1$, be the disk chain complex with $R$ in degrees $n,n-1$ and identity differential. The generating cofibrations for nonnegative chain complexes areEach is a model cofibration: its degreewise cokernel is a copy of $R$ in one degree. The extra degree-zero generator must not be omitted.

Here is the lifting characterization of an acyclic chain-complex fibration. A right lifting property for $S^{n-1}(R)\to D^n(R)$ says that whenever $x\in Z_{n-1}X$ and $y\in Y_n$ satisfy $f(x)=dy$, there is $z\in X_n$ with $dz=x$ and $f(z)=y$. The generator $0\to S^0(R)$ says that $f_0$ is surjective. For a degreewise surjective quasi-isomorphism, its kernel is acyclic by the long exact sequence in homology. Choose a preliminary lift $z_0$ of $y$. Then $d{z}_0-x$ is a cycle in the kernel; subtract an element of the kernel whose boundary is this cycle. This gives the required $z$.

Conversely the lifting conditions make the kernel acyclic: take $y=0$ and any kernel cycle $x$. They also imply degreewise surjectivity inductively. Having surjectivity below degree $n$, lift $dy$ to an element of $X_{n-1}$, correct its boundary within the acyclic kernel to make it a cycle, and apply the lifting condition to lift $y$. For $n=1$ the chosen degree-zero element is already a cycle. A degreewise surjective map with acyclic kernel is a quasi-isomorphism by the long exact sequence in homology. Hence $I$-injectives are exactly acyclic fibrations.

An object $A$ is sequentially small if for every sequential diagram $Z_0\to Z_1\to \cdots$, the natural mapis bijective. Surjectivity says that a map out of $A$ factors through a finite stage; injectivity says that two such maps agreeing in the colimit agree at a later finite stage. The relative version restricts to the stated class of sequential diagrams. The domains in $I$ have this property: maps from $S^{n-1}(R)$ are cycles in one degree, and filtered colimits commute with these finite equations. Their sequential smallness is also permitted by the hint.

Apply the small object argument to the map $a:X\to Y$. Put $Z_0=X$, $p_0=a$. Given $p_k:Z_k\to Y$, take the set of all commutative squares with a generator $i:A_i\to B_i$ on the left and $p_k$ on the right. Form the pushout of the coproduct of all these $i$ along their maps into $Z_k$, obtaining $Z_{k+1}$. Each bottom map $B_i\to Y$ induces the compatible map $p_{k+1}:Z_{k+1}\to Y$. Repeat for all $k\ge 0$, and set $Z_{\infty }={\mathrm{colim}}_k{Z}_k$. This givesConcretely, attaching $S^{n-1}(R)\to D^n(R)$ adds a free generator in degree $n$ whose boundary is the chosen existing cycle; attaching $0\to S^0(R)$ adds a free degree-zero generator. Thus $j$ is an injection, and its cokernel in each degree is free on the newly attached generators. It is therefore a model cofibration; this also illustrates the relative cell complex description of the left factor.

For a lifting square into $Z_{\infty }\to Y$, sequential smallness of its domain $A_i$ factors the top map through some $Z_k$. The commutativity is already an equality in $Y$, so this is one of the squares attached at stage $k+1$. Its new cell supplies a lift $B_i\to Z_{k+1}\to Z_{\infty }$. Hence $p$ has the right lifting property with respect to every element of $I$, and is an projective acyclic fibration. Every map has the required cofibration–acyclic-fibration factorization. Only the domains need sequential smallness; there is no requirement that $X$, $Y$ or the coproduct of all cells be small.

Pushing an inclusion $i:A\to B$ along $f:A\to A\prime$ gives a new module extension with kernel $A\prime$ and the same quotient. This pushout realizes the covariance of the degree-one Ext functor in its second argument.

For a natural transformation $a:X\to Y$ of set-valued functors, form the pushout by taking two copies of each $Y(C)$ and identifying the two copies of every element in the image of $a_C$. Naturality makes all induced maps well-defined. If $y\in Y(C)$ is outside that image, its two copies remain distinct. Thus the two maps $Y\rightrightarrows Y{⨿}_{X}Y$ agree after $a$ but differ at $y$, proving $a$ is not an epimorphism. Conversely, componentwise surjective functions allow cancellation at every object, proving that $a$ is an epimorphism. This argument does not assume a natural choice of preimages.

For a frame homomorphism $h:B\to A$, nuclei are equivalent when their restrictions to $h(B)$ agree. The least representative of the class of $j$ isThe least representatives form a frame generated by $c(A)$ and $o(h(B))$. To see minimality, each summand is below $j$ and its value at $hb$ is $j(hb)$; any equivalent nucleus therefore bounds every summand. This construction gives the pushout of $B\to N(B)$ along $h$.

In a coherent category, maps $A\to Y$ and $B\to Y$ agreeing on the intersection define a relation from $A\cup B$ to $Y$ by taking the union of their graphs. It is total because the two subobjects cover their union, and single-valued because the maps agree on overlap. Total functional relations define morphisms, giving a unique glued map. This is the pushout universal property without assuming a general supply of pushouts.