Seifert-van Kampen theorem

For a suitably path-connected open cover $X=U\cup V$, the fundamental group of $X$ is the pushout of the homomorphisms from ${\pi }_1(U\cap V)$ to ${\pi }_1(U)$ and ${\pi }_1(V)$.

Attach a closed interval to a sphere $S^2$ at two distinct endpoints, with its interior disjoint from the sphere. The fundamental group is infinite cyclic group. Remove a point of the sphere away from the endpoints: the remaining plane has a deformation retraction onto an arc between the endpoints, keeping the added closed interval fixed. This gives a circle. An open cap about the removed point is simply connected, and its punctured overlap has trivial map into the first fundamental group because it lies in the plane. The Seifert-van Kampen theorem therefore adds no further relator.

Forthe Seifert-van Kampen theorem gives

Given homomorphisms $i_A:C\to A$ and $i_B:C\to B$, the amalgamated free product $A{\ast }_{C}B$ comes with homomorphisms from $A$ and $B$ that agree on $C$. It is universal with this property: every compatible pair of homomorphisms $A\to H$ and $B\to H$ factors uniquely through $A{\ast }_{C}B$.

The free product $A\ast B$ is the amalgamated free product over the trivial group. Its nonidentity elements have unique reduced alternating normal forms.

Let $G=⟨x,y\mid y^2=1⟩$. For each $n\ge 2$, a free basis of a subgroup of rank $n$ isA covering-space proof uses $r=n-1$ spheres double-covering the real projective plane, with antipodal points $a_j,b_j$ over the wedge point. Add lifted $x$ edges in a cyclic order through the $a_j$, and an $x$ loop at each $b_j$. This is a connected $2r$-sheeted cover of a circle wedged with a projective plane. After collapsing one connecting arc in each simply connected sphere, its fundamental group is that of a graph with $r$ vertices and $2r$ edges, hence free of rank $r+1$.

A spanning tree along the first $r-1$ cyclic edges gives the displayed words: the closing cycle is $x^r$, and the loop reached through $a_j,b_j$ is $x^{j}yxy{x}^{-j}$. Injectivity of the induced fundamental-group map of a covering map proves they form a free basis in $G$.

Suppose $P$ has a soluble word problem for a group and $z$ generates an infinite cyclic group. In $P\ast ⟨z⟩$, for $g\ne 1$ the words $gz$ and $[g,z]$ are cyclically reduced with two and four syllables respectively. Their nonzero powers have syllable lengths $2|k|$ and $4|k|$. Given an input's reduced free product form, its length bounds the possible exponents; compare the finitely many candidates using the base word algorithm. Membership and the exponent in each of these infinite cyclic subgroups are therefore computable, with no algorithm for the order of $g$ required.

A subgroup $A$ is a free factor of $G$ if $G$ is isomorphic to a free product $A\ast B$ by an isomorphism restricting to the inclusion of $A$, for some subgroup $B$. The normal form theorem for a free product ensures that both factors embed. A free factor need not be a normal subgroup.

Given group homomorphisms $f_A:A\to H$ and $f_B:B\to H$, there is exactly one group homomorphism $A\ast B\to H$ extending both. Its value on an alternating normal form is the product of the factor-map values. Thus the free product is the coproduct in the category of groups.

A free product $A\ast B$ is nontrivial as a splitting when both factors differ from the trivial group. Alternating words $(ab{)}^r$ with $a,b\ne 1$ in different factors show that every such splitting gives an infinite group. The distinct normal forms $ab$ and $ba$ also show that the group is not an abelian group.

For nontrivial groups $A,B$, the Cayley graph of $A\ast B$ with $S=(A\setminus \{1\})\cup (B\setminus \{1\})$ has the edges of the Bass-Serre tree as its vertices. Two such vertices are adjacent exactly when the tree edges share an endpoint. Mapping each group element to its tree-edge midpoint is an isometry on the vertex metrics and has image within distance $1/2$ of every tree point, hence gives a quasi-isometry of metric graphs. The normal form theorem for a free product identifies $|g{|}_S$ with reduced syllable length. The factor subgroups have diameter one in this metric.

A normal form for free groups and free product of groups represents elements by reduced words. In a free group, adjacent inverse letters cancel; in a free product, alternating nonidentity syllables from different factors are unique. In an amalgamated free product, chosen coset representatives produce the corresponding reduced form relative to the common subgroup.

Every element of a free product $G\ast H$ has a unique expression $u_1\cdots u_n$ in which each $u_i$ is a nonidentity element of one factor and adjacent syllables belong to different factors. The identity corresponds to the empty word.

A syllable is a nonidentity element of one factor in a free product word. A reduced normal form has successive syllables in different factors; its syllable length is the number of these elements. This length records factor elements, rather than letters in a chosen generating set.

If $X=U_1\cup U_2$, the sets $U_1,U_2$, and their intersection are path-connected, and the base point lies in the intersection, then ${\pi }_1(X)$ is generated by the images of ${\pi }_1(U_1)$ and ${\pi }_1(U_2)$. Subdivide each loop into arcs lying in cover members and join every transition point to the base point within the intersection; the inserted paths telescope.

A Möbius band retracts to its core circle, and its boundary wraps twice around the core. Attaching it to $X$ along a loop representing $w\in {\pi }_1(X)$ therefore adjoins a generator $x$ with relation $x^2=w$.

Glue the boundary of a Möbius band to the projective line in the real projective plane by a mapping degree $n$. The band is the mapping cylinder of the degree-two boundary covering of its core. Collapsing the connecting tree in the resulting double mapping cylinder gives a CW complex with one vertex, edges $a,b$, and two attaching words $a^2$ and $a^{-n}{b}^2$. Its cellular chain complex has degree-two boundary matrix $\left(\begin{array}{cc}2& -n\\ 0& 2\end{array}\right)$. Its determinant is four, so $H_2=0$. The Smith normal form gives $H_1\cong \mathbb{Z}/4$ for odd $n$ and $(\mathbb{Z}/2{)}^2$ for even $n$; $H_0=\mathbb{Z}$ and higher homology vanishes. The space cannot be homotopy equivalent to a closed manifold: mod-two homology forces a potential manifold to be a surface, and its Euler characteristic is one, whereas the only such closed connected surface is the projective plane, with different integral $H_1$.

Attach Möbius bands to a torus along the classes $ab$ and $a^2{b}^3$. If their core generators are $x,y$, the resulting presentation isBecause the exponent vectors $(1,1)$ and $(2,3)$ form a unimodular basis of ${\mathbb{Z}}^2$, eliminating $a,b$ givesThis group maps onto $S_3$ by $x\mapsto (12)$ and $y\mapsto (23)$, so it is nonabelian.

Attaching a disc to $X$ along a based loop $\alpha$ kills precisely the normal closure of its homotopy class:

A group presentation gives a two-dimensional cell complex with one vertex, one oriented loop for each generator, and one 2-cell attached along each relator. Its fundamental group is the presented group.

Attach three 2-cells to $S^1\vee S^1$ along loops representing $a^2$, $b^3$, and $(ab{)}^2$. The resulting group isIndeed, $a\mapsto (12)$ and $b\mapsto (123)$ gives a surjection to $S_3$, while $aba=b^{-1}$ reduces every word to one of $b^i$ or $a{b}^i$, so the presented group has at most six elements.