 Solution

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.