The Seifert–van Kampen Theorem

By [F5], ${\alpha }^{-1}(U)$ and ${\alpha }^{-1}(V)$ form an open cover of the compact metric interval $I$. Choose a Lebesgue number $\delta >0$ by [F1], then choose $q\ge 1$ with $1\mathrm{/}q<\delta$ by [F7] and use the subdivision $t_i=i\mathrm{/}q$. Each restricted path ${\alpha }_i:=\alpha {\mathrm{\mid }}_{[t_{i-1},t_i]}$, reparametrized to $I$, is continuous by [F8] and lies wholly in $U$ or wholly in $V$; assign it to $U$ if its image lies in $U$, and to $V$ otherwise. Merge adjacent pieces assigned to the same set. After merging, every interior subdivision value $\alpha (t_i)$ lies in $U\cap V$; the construction also admits the constant loop and the case of one retained piece.

Put ${\lambda }_0={\lambda }_m=c_{x_0}$. For each remaining interior vertex, [F2] makes the family of paths in $U\cap V$ from $x_0$ to $\alpha (t_i)$ nonempty, so [F4] supplies paths ${\lambda }_i$ for the finitely many vertices. If ${\alpha }_i$ lies in $A_i\in \{U,V\}$, then ${\beta }_i:={\lambda }_{i-1}\ast {\alpha }_i\ast {\bar{\lambda }}_i$ is a based loop in $A_i$.

In the product $[{\beta }_1]\cdots [{\beta }_m]$, every adjacent pair ${\bar{\lambda }}_i\ast {\lambda }_i$ cancels up to endpoint-fixed path homotopy, and the outside connectors are constant. Thus [F3] gives $[\alpha ]=(j_{A_1}{)}_{\ast }[{\beta }_1]\cdots (j_{A_m}{)}_{\ast }[{\beta }_m]$. For $m=1$ this is the single factor $[\alpha ]$, and for a constant loop it is the identity, so every loop class has the asserted factorization.

Delete the $j$-th coordinate to identify $e_j^{\perp }$ with ${\mathbb{R}}^n$, and write $z=u+t{e}_j$. Projection from $e_j$ is ${\sigma }_{+}(z)=u\mathrm{/}(1-t)$ on $U_j$, with inverse ${\tau }_{+}(y)=\frac{2y}{\parallel y{\parallel }_2^2+1}+\frac{\parallel y{\parallel }_2^2-1}{\parallel y{\parallel }_2^2+1}{e}_j\mathrm{.}$ Projection from $-e_j$ is ${\sigma }_{-}(z)=u\mathrm{/}(1+t)$ on $V_j$, with inverse ${\tau }_{-}(y)=\frac{2y}{\parallel y{\parallel }_2^2+1}+\frac{1-\parallel y{\parallel }_2^2}{\parallel y{\parallel }_2^2+1}{e}_j\mathrm{.}$ Since $\parallel y{\parallel }_2^2\ge 0$, the inverse denominator $\parallel y{\parallel }_2^2+1$ is positive. The sphere equation $\parallel u{\parallel }_2^2+t^2=1$ and deletion of the relevant pole give $1-t\mathrm{\ne }0$ on $U_j$ and $1+t\mathrm{\ne }0$ on $V_j$. The same equation shows that both inverse formulas land on $S^n$, and direct substitution gives ${\sigma }_{\pm }{\tau }_{\pm }=\mathrm{id}$ and ${\tau }_{\pm }{\sigma }_{\pm }=\mathrm{id}$.

The coordinate $t$ is continuous, so $U_j=\{z\in S^n:t<1\}$ and $V_j=\{z\in S^n:t>-1\}$ are open by [F4]. They cover the sphere because no point has both $t=1$ and $t=-1$. The formulas in step 1.1 are continuous by [F2] and [F3], so they are the asserted homeomorphisms. Moreover ${\sigma }_{+}$ sends the deleted point $-e_j$ to $0$, hence restricts to a homeomorphism $U_j\cap V_j\cong {\mathbb{R}}^n\setminus \{0\}$.

The space ${\mathbb{R}}^n$ is nonempty and convex, so [F5] makes it simply connected. Each homeomorphism in step 2.1 transports paths, and [F7] applied to it and its inverse transports the one-element fundamental group at every basepoint. Thus [F8] makes both $U_j$ and $V_j$ simply connected.

By [F6] and the last homeomorphism in step 2.1, $U_j\cap V_j$ is path-connected. This proves every assertion.

The point $x_0$ cannot be a pole on both the first and second coordinate axes. Choose the first axis unless $x_0$ is one of its two poles, and choose the second axis otherwise. For that axis, [L1] supplies an antipodal open cover $S^n=U\cup V$ with $x_0\in U\cap V$, with $U$ and $V$ simply connected and $U\cap V$ path-connected. Any two points of $S^n$ can be joined by going inside their respective cover members to a point of the nonempty overlap and then inside the overlap, so $S^n$ is path-connected.

By [L2], every element of ${\pi }_1(S^n,x_0)$ is a product of classes induced from ${\pi }_1(U,x_0)$ and ${\pi }_1(V,x_0)$. Both groups have one element by [L1], so every factor, and hence every such product, is the identity. Thus ${\pi }_1(S^n,x_0)$ has one element.

The sphere is nonempty, step 1.1 makes it path-connected, and step 2.1 applies to the arbitrary basepoint $x_0$. Therefore every basepoint has a one-element fundamental group, so [F1] makes $S^n$ simply connected.

By [F4], $H^{-1}(U)$ and $H^{-1}(V)$ form an open cover of the compact metric square $I^2$. By [F1] and [F2], choose a Lebesgue number $\delta >0$ for this cover in $d_{\mathrm{\infty }}$.

Take $q\ge 1$ with $1\mathrm{/}q<\delta$, using [F5]. Start with the uniform cuts $k\mathrm{/}q$ in each coordinate, adjoin the finitely many prescribed bottom and top cuts to the horizontal list, and sort each resulting finite set after removing repetitions. Every gap in either partition is at most $1\mathrm{/}q<\delta$, and both lists contain $0$ and $1$.

Each closed grid rectangle is nonempty and has $d_{\mathrm{\infty }}$-diameter at most $1\mathrm{/}q<\delta$, so [F2] places it inside $H^{-1}(U)$ or $H^{-1}(V)$. Its image therefore lies in the corresponding cover member, and the constructed horizontal partition refines both prescribed boundary subdivisions.

The value of a subordinate factorization $[\alpha ]=(j_{A_1}{)}_{\ast }[a_1]\cdots (j_{A_m}{)}_{\ast }[a_m]$ is, by definition, $i_{A_1}[a_1]\cdots i_{A_m}[a_m]\in P$. Subdividing a factor only replaces one factor by a product in the same fundamental group. If a connector at a subdivision point is changed, the two connectors differ by a loop in $U\cap V$, and [F1] gives the same element whether that correction is read in the $U$ factor or the $V$ factor. Thus refinements and connector changes preserve the value.

Choose an endpoint-fixed path homotopy $H$ from $\alpha$ to $\beta$. By [L2], take a subordinate rectangular grid refining the prescribed subdivisions of both factorizations. At every grid vertex, choose a path to $x_0$ inside $U$ if all adjacent rectangles are assigned to $U$, inside $V$ if all are assigned to $V$, and inside $U\cap V$ if both assignments occur. Each required path family is nonempty by path-connectedness, and [F2] licenses the finite selection; on the bottom and top edges use the prescribed connectors after the harmless adjustments of step 1.1.

For an oriented grid edge, concatenate its chosen endpoint connectors with the image of the edge. This gives a based loop in either adjacent cover member; if the adjacent assignments differ, both connectors lie in $U\cap V$, so [F1] identifies the two readings in $P$. Around one grid rectangle, the four edge loops multiply to the identity because the restriction of $H$ to that rectangle contracts its boundary inside its assigned cover member. Multiplying these boundary identities row by row cancels every interior edge with its reverse, leaving exactly the refined bottom word and the inverse of the refined top word. Hence the two words have equal value in $P$.

Step 1.1 identifies the refined boundary words with the values of the original factorizations, while step 3.1 identifies those refined values with each other. Therefore every factorization of $\alpha$ and every factorization of $\beta$ have the same value in $P$.

Choose the pushout $P$ supplied by [L3]. The two inclusion-induced homomorphisms from ${\pi }_1(U,x_0)$ and ${\pi }_1(V,x_0)$ to ${\pi }_1(X,x_0)$ agree on ${\pi }_1(U\cap V,x_0)$, since both composites are induced by the same inclusion. By [F1], they therefore define a unique homomorphism $\mathrm{\Phi }:P\to {\pi }_1(X,x_0)$ satisfying $\mathrm{\Phi }{i}_U=(j_U{)}_{\ast }$ and $\mathrm{\Phi }{i}_V=(j_V{)}_{\ast }$.

For $c\in {\pi }_1(X,x_0)$, let $\mathrm{\Psi }(c)$ be the element of $P$ that is the value of some subordinate factorization of some loop representing $c$. Such a factorization exists by [L1], and [L2] says that all representatives and all their factorizations give the same value. Thus exactly one such element exists for each $c$, so this rule defines a function without selecting factorizations globally. Concatenating two factorizations concatenates their words, hence $\mathrm{\Psi }$ is a homomorphism.

If $c$ is represented by a factorized loop, applying $\mathrm{\Phi }$ to its pushout word restores the same product of inclusion-images, which is $c$ by [L1]. Hence $\mathrm{\Phi }\mathrm{\Psi }={\mathrm{id}}_{{\pi }_1(X,x_0)}$.

For $a\in {\pi }_1(U,x_0)$, the one-factor factorization of its image gives $\mathrm{\Psi }((j_U{)}_{\ast }a)=i_U(a)$, and similarly on the $V$ factor. Thus $\mathrm{\Psi }\mathrm{\Phi }:P\to P$ agrees with the identity after composition with both factor maps. Uniqueness in [F1] makes $\mathrm{\Psi }\mathrm{\Phi }={\mathrm{id}}_P$.

The homomorphisms $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse, so $\mathrm{\Phi }$ is an isomorphism and ${\pi }_1(X,x_0)$ has the asserted pushout property.

By [F2], ${\pi }_1(U\cap V,x_0)$ is the trivial group, so the two maps in the pushout of [L1] are the unique homomorphisms from the trivial group.

The two maps from the trivial group are injective, so [F1] identifies their pushout with ${\pi }_1(U,x_0)\ast {\pi }_1(V,x_0)$. Combining this with [L1] gives the displayed isomorphism.

By [F4], ${\pi }_1(V,x_0)$ is trivial. Thus [L1] identifies ${\pi }_1(X,x_0)$ with the pushout of $k_{\ast }$ and the unique homomorphism from ${\pi }_1(U\cap V,x_0)$ to the trivial group.

By [F3], the free product of ${\pi }_1(U,x_0)$ with the trivial group is canonically ${\pi }_1(U,x_0)$. Under this identification, [F1] says that the pushout in step 1.1 is ${\pi }_1(U,x_0)/⟨\text{\hspace{0.17em}⁣}⟨\mathrm{im}{k}_{\ast }⟩\text{\hspace{0.17em}⁣}⟩\mathrm{.}$

The canonical map from ${\pi }_1(U,x_0)$ to this quotient is exactly $(j_U{)}_{\ast }$ under [L1]. A quotient map is surjective and has the quotienting normal subgroup as its kernel, so the asserted surjectivity and kernel formula follow.

The tagged quotient for $W_r\vee Q$ differs from that for $W_{r+1}$ only by grouping the old tagged summands; the quotient maps in both directions preserve every tag and are continuous by [F2], so they are inverse homeomorphisms. Under this identification, define $A_r$ to contain all of the old wedge $W_r$ and the arc $O$ in the new circle. Define $B_r$ to contain the whole new circle and the arc $O$ in every old circle. Their inverse images under the wedge quotient are open, by [F1] and the disjoint-union topology, and each inverse image is saturated because every listed arc contains its basepoint. Hence $A_r$ and $B_r$ are open; they cover $W_{r+1}$, and their intersection consists exactly of one copy of $O$ in every circle, with all their basepoints identified.

Use the coordinate $t\in (-1\mathrm{/}4,1\mathrm{/}4)$ supplied by [F1] and the contraction $t\mapsto (1-s)t$. On $A_r$, contract only the new-circle arc and fix $W_r$; on $B_r$, contract every old-circle arc and fix the new circle. The formulas agree at the tagged basepoints and are jointly continuous away from the wedge point. At the wedge point, a target neighbourhood contains an arc $\mathrm{\mid }t\mathrm{\mid }<{\epsilon }_j$ in every incident branch; there are finitely many branches, so their minimum is positive, and the contraction never increases $\mathrm{\mid }t\mathrm{\mid }$. Together with the fixed trace on the retract, this gives a product neighbourhood mapped into the target neighbourhood, proving joint continuity there. Thus the formulas are deformation retractions as in [F3], and [F5] makes the two retract inclusions induce fundamental-group isomorphisms.

Apply the same contraction simultaneously on every arc of $A_r\cap B_r$. Joint continuity away from $w$ is coordinatewise, and at $w$ the same finite-minimum neighbourhood argument from step 2.1 applies to all incident arcs. This gives a deformation retraction of the overlap onto $w$. The construction also covers $r=0$, when $W_0$ is the point $w$, and $r=1$, when the old wedge has one circle.

Each point of any of the three sets can be joined within its circle arc or circle to $w$, so [F4] makes all three path-connected. At the basepoint $w$, step 3.1 and [F5] identify the overlap fundamental group with that of a point, hence with the one-element group. For any other basepoint $y$, a path $\rho$ from $w$ to $y$ gives an isomorphism from the group at $w$ to the group at $y$ by $[\alpha ]\mapsto [\bar{\rho }\ast \alpha \ast \rho ]$, with reverse-path inverse, using [F7]; hence its fundamental group is one-element as well. The overlap is nonempty, so [F6] makes it simply connected.

The empty wedge $W_0$ is a point by definition. Every based loop in a point is constant, so ${\pi }_1(W_0,w)$ is the one-element group, which is the free group on the empty basis and has rank $0$.

The group of one circle is infinite cyclic by [F1], so its standard loop is a one-element free basis. This is the first successor case and fixes the basis convention used below.

Assume ${\pi }_1(W_r,w)$ is free on the $r$ standard circle loops. By [L1] and [L2], the successor wedge satisfies ${\pi }_1(W_{r+1},w)\cong {\pi }_1(W_r,w)\ast {\pi }_1(Q,[0])\mathrm{.}$

The induction hypothesis and [F1] identify the two factors as free groups on disjoint bases consisting of the old $r$ standard loops and the new standard loop. By [F2], their free product is free on the union, exactly the $r+1$ standard loops of $W_{r+1}$.

Step 1.1 is the base case and steps 1.3 and 2.1 prove the successor implication, so [F3] gives the result for every $r\in \mathbb{N}$. The basis has $r$ elements, hence the rank is $r$ by definition.

Apply [L1] with $r=2$ to obtain that ${\pi }_1(W_2,w)$ is free on its standard circle loops.

Those standard loops are precisely $a$ and $b$, so ${\pi }_1(W_2,w)\cong F(a,b)$.

Projection sends a loop in $X\times Y$ to one loop in each factor. If two product loops are path-homotopic relative to endpoints, composing the homotopy with either projection gives a path homotopy of the coordinate loops by [F4]. Thus the displayed rule $\mathrm{\Theta }$ is well defined.

Conversely, [F1] makes $s\mapsto (\alpha (s),\beta (s))$ a continuous based loop. Pairing two endpoint-fixed homotopies gives a product homotopy by the same characteristic property, so the resulting class depends only on $[\alpha ]$ and $[\beta ]$. This defines a function $\mathrm{\Xi }$ from the direct product to ${\pi }_1(X\times Y,(x_0,y_0))$.

Projections commute with concatenation, so [F2] and [F3] show that $\mathrm{\Theta }$ is a group homomorphism.

Coordinatewise concatenation shows that $\mathrm{\Xi }$ is a homomorphism. By construction $\mathrm{\Theta }\mathrm{\Xi }([\alpha ],[\beta ])=([\alpha ],[\beta ])$, and uniqueness of a map with given coordinates in [F1] gives $\mathrm{\Xi }\mathrm{\Theta }[\gamma ]=[\gamma ]$. Hence $\mathrm{\Theta }$ and $\mathrm{\Xi }$ are inverse group isomorphisms.

Let $f:(X,x_0)\to (X^{\mathrm{\prime }},x_0^{\mathrm{\prime }})$ and $g:(Y,y_0)\to (Y^{\mathrm{\prime }},y_0^{\mathrm{\prime }})$ be pointed continuous maps. By [F1], their product $f\times g$ is pointed and continuous. Composition with any pointed continuous map preserves endpoint-fixed homotopies and commutes with loop concatenation, so [F2], [F4], and [F5] make all three induced maps in the naturality square well-defined homomorphisms. For every product-loop class $[\gamma ]$, both $(f_{\ast }\times g_{\ast }){\mathrm{\Theta }}_{X,Y}[\gamma ]$ and ${\mathrm{\Theta }}_{X^{\mathrm{\prime }},Y^{\mathrm{\prime }}}(f\times g{)}_{\ast }[\gamma ]$ equal $([f\circ {\pi }_X\circ \gamma ],[g\circ {\pi }_Y\circ \gamma ])\mathrm{.}$ Thus the naturality square commutes for every pair $(f,g)$, so the displayed group isomorphism is natural.

By the torus definition and [L1], ${\pi }_1(T^2,([0],[0]))\cong {\pi }_1(\mathbb{R}\mathrm{/}\mathbb{Z},[0])\times {\pi }_1(\mathbb{R}\mathrm{/}\mathbb{Z},[0])\mathrm{.}$

Applying [F1] in both coordinates gives the displayed isomorphism with $\mathbb{Z}\times \mathbb{Z}$.