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In a free group, a reduced word has no adjacent inverse letters. Successively cancelling such pairs gives the unique reduced representative in the chosen free generators. In a free product, a reduced word instead uses nonidentity factor elements as syllables in a free product, with consecutive syllables in a free product from different factors; the alternating syllables in a free product are unique.

For an amalgamated free product $A{\ast }_{C}B$, the usual reduced form relative to a common subgroup requires the maps $C\to A$ and $C\to B$ to be injective. Chosen representatives of cosets then distinguish the alternating factors and the remaining common-subgroup element. The exact form depends on the chosen left/right representative convention.

For general group pushouts defined by arbitrary homomorphisms, factors can collapse and this embedded-subgroup normal form is unavailable. For example, take $A=B=C=\mathbb{Z}$, map $C$ identically into $A$, and map it trivially into $B$. The pushout kills all of $A$ and is isomorphic to $B$, so words in nonidentity elements of $A$ cannot retain the usual nontrivial-factor normal form.

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.

A canonical presentation of a group uses a symbol $[g]$ for each $g\in G$ and relations $[g][h]=[gh]$ for every pair. Evaluation to $G$ is surjective, and $g\mapsto [g]$ is its inverse homomorphism; the relation with $g=h=1$ implies $[1]=1$. Thus this presentation recovers any group.

The pushout of the two homomorphisms isHere $\ast$ is the free product, and the denominator is the normal closure. Its canonical maps $j_i:G_i\to G$ satisfy $j_1{\varphi }_1=j_2{\varphi }_2$. The universal property says that maps $u_i:G_i\to H$ agreeing on $G_0$ factor uniquely as $u_i=u{j}_i$ for one homomorphism $u:G\to H$. The free-product universal property first gives the map before quotienting, and agreement makes it kill precisely the imposed relations, proving this property.

Suppose $G_0$ is generated by $s_1,\dots ,s_r$, and $G_i=⟨S_i\mid R_i⟩$ are finite presentations with disjoint generator sets. Choose words $w_{ij}$ representing ${\varphi }_i(s_j)$. ThenAgreement on these generators implies agreement on every word in $G_0$, so these finitely many extra relations suffice by the universal property. No finite presentation for $G_0$ is required.

The Seifert-van Kampen theorem states that for a space covered by path-connected open sets $U,V$ with path-connected nonempty intersection and basepoint in that intersection, ${\pi }_1(U\cup V)$ is the pushout of ${\pi }_1(U)\leftarrow {\pi }_1(U\cap V)\to {\pi }_1(V)$.

To see the two bands in the Klein bottle model, take one region to be $1/4\le y\le 3/4$ and the other to be the outer strips $0\le y\le 1/4$, $3/4\le y\le 1$, joined along the top-bottom identification. Each is an interval bundle over the horizontal circle, and going once around reverses its transverse coordinate. For the middle band use the coordinate $y-1/2$; for the outer band use the signed coordinate about zero modulo one. Both are therefore Möbius bands. Their common boundary is the one circle represented by the lines $y=1/4,3/4$, joined at the identified vertical sides. They cover the whole bottle and are glued along that circle.

A Möbius band deformation retracts onto its core circle, so its fundamental group is infinite cyclic. Going once around its boundary traverses the core twice: one pass along one boundary edge changes sides, and the second pass returns. After choosing core generators and consistent boundary orientation, the two inclusion maps from the common circle send its generator to $x^2$ and $y^2$. Thicken the bands slightly so their overlap is an annular neighbourhood of their common boundary. Seifert-van Kampen theorem then givesReversing a chosen generator absorbs an opposite boundary parametrization, so the presentation is independent of that sign choice.

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 $h$.

Let $X=U\cup V$ be an open cover such that $U$, $V$ and $U\cap V$ are nonempty and path-connected, and choose a base point $x_0\in U\cap V$. The inclusions induce a pushout of groups:The maps from the overlap need not be injective. The amalgamated free product here means the universal group receiving the two inclusion-induced homomorphisms with their overlap relations.

A loop can be subdivided into pieces in $U$ or $V$; connecting their endpoints through the path-connected overlap expresses its class using the two local fundamental groups. Subdividing a homotopy shows that the overlap relations are exactly the identifications needed.

The overlap condition matters. Two open arcs can cover a circle while their intersection has two components. Both arcs have trivial fundamental group, but the circle has fundamental group $\mathbb{Z}$. The displayed single-basepoint formula cannot be applied to that disconnected overlap; a groupoid version keeps the extra component information.

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.