The Loop-Subdivision Proof of Van Kampen

From Local Loops to a Global Loop

The Seifert–van Kampen theorem describes loops in a union of two open sets using loops in each set and the identifications made on their intersection. The point requiring proof is that this local description captures not only every loop, but also every homotopy between loops. Subdivision supplies both facts: compact paths break into pieces in the cover sets, and compact homotopies break into small cells in the cover sets.

Throughout, suppose $X=U\cup V$, where $U,V$ are open subsets of $X$. Assume $U$, $V$, and $U\cap V$ are path-connected, and choose a basepoint $x_0\in U\cap V$. The fundamental group $\pi_1(Y,x_0)$ consists of based loops in $Y$, up to homotopy that keeps the basepoint fixed; its operation is concatenation. The inclusion maps from $U\cap V$ into $U$ and $V$ induce homomorphisms on fundamental groups.

Theorem (Seifert–van Kampen, two-set form): Under these hypotheses, the homomorphisms induced by inclusion give an isomorphism $$ \pi_1(X,x_0)\cong \bigl(\pi_1(U,x_0)*\pi_1(V,x_0)\bigr)\big/N, $$ where $N$ is the smallest normal subgroup containing $i_U(c)i_V(c)^{-1}$ for every $c\in\pi_1(U\cap V,x_0)$. Here $i_U$ and $i_V$ are induced by the two inclusions. Equivalently, $\pi_1(X,x_0)$ is the pushout of the two inclusion-induced homomorphisms.

The free product in this statement allows words alternating between loops from $U$ and $V$. The quotient imposes precisely the identifications required for a loop in the intersection to count as the same loop in either set. We prove the result by first establishing the corresponding statement for paths with arbitrary endpoints, and then returning to based loops.

Subdividing Paths and Homotopies

Lemma (path subdivision): If $\alpha:[0,1]\to X$ is a path and $X=U\cup V$ with $U,V$ open, then there is a finite subdivision $0=t_0<t_1<\cdots<t_n=1$ such that each restricted path $\alpha|_{[t_{j-1},t_j]}$ has image in $U$ or in $V$.

Proof. The inverse images $\alpha^{-1}(U)$ and $\alpha^{-1}(V)$ are open in $[0,1]$ and cover that compact interval. A finite open cover of a compact metric space has a Lebesgue number: some $\delta>0$ is such that every subset of diameter less than $\delta$ lies in one member of the cover. To recall why, choose for each point a ball of radius $r>0$ contained in a member of the cover. Finitely many balls with half those radii cover the space by compactness. The minimum of their half-radii is positive; any set of smaller diameter that meets one of the half-radius balls lies in the corresponding full-radius ball. Now subdivide $[0,1]$ into intervals of length less than $\delta$. Each interval lies in one of the inverse images, which proves the claim. $\square$

The same compactness argument applies to a homotopy $H:[0,1]^2\to X$. The inverse images of $U,V$ cover the square. Choose a Lebesgue number for this cover and a rectangular grid fine enough that every closed cell has diameter less than that number. Then the image of each cell lies wholly in $U$ or wholly in $V$. This is the key reason the proof handles homotopies, rather than just individual paths.

Worked Example: Subdividing a Path

Let $X=(-2,2)$ with the subspace topology from $\mathbb{R}$, let $U=(-2,1)$, and let $V=(0,2)$. These are open in $X$, and $U\cup V=X$. Consider the path $\alpha(t)=-1+2t$, for $0\le t\le1$, from $-1$ to $1$.

Use the subdivision $t_0=0,\ t_1=3/5,\ t_2=1$. On the first subinterval, $\alpha([0,3/5])=[-1,1/5]$. Since $-1>-2$ and $1/5<1$, this whole image lies in $U$. On the second, $\alpha([3/5,1])=[1/5,1]$. Since $1/5>0$ and $1<2$, this whole image lies in $V$. The transition point is $\alpha(3/5)=1/5$, which belongs to $U\cap V=(0,1)$.

Check: the two restricted paths concatenate to the original path because their parameter intervals are $[0,3/5]$ and $[3/5,1]$, meeting at the same value $1/5$. The subdivision does not turn either piece into a based loop; it gives a factorization of a path into paths with intermediate endpoints.

Worked Example: A Grid for a Homotopy

Give $X=[0,1]^2$ the subspace topology from $\mathbb{R}^2$, and set $U=\{(x,y)\in X:x<3/5\}$ and $V=\{(x,y)\in X:x>2/5\}$. These are open in $X$, cover $X$, and overlap where $2/5<x<3/5$. Consider the homotopy $H(s,t)=(s,t)$, which maps the parameter square onto $X$.

Divide each parameter coordinate at $0,1/2,1$, making four closed rectangular cells. On a cell with $0\le s\le1/2$, the first coordinate of $H(s,t)$ satisfies $s\le1/2<3/5$, so the image cell lies in $U$. On a cell with $1/2\le s\le1$, it satisfies $s\ge1/2>2/5$, so the image cell lies in $V$. Thus every cell lies in a cover set, even though the whole image square need not lie in either one.

The shared vertical edges have $s=1/2$, hence lie in $U\cap V$. Each cell boundary is null-homotopic within its assigned set: the cell image is a rectangle, and straight-line contraction to any point of that rectangle stays inside it. This grid illustrates how local boundary relations can be assembled into a relation on the boundary of the whole square.

The Groupoid Subdivision Argument

A groupoid is like a group with multiple objects: its morphisms are path-homotopy classes of paths, and its composition is concatenation when endpoints match. Write $\Pi(Y)$ for the groupoid whose objects are points of $Y$ and whose morphisms are paths in $Y$ up to homotopy fixing both endpoints. Use the objects of $X$ throughout. Thus $\Pi(U)$ has objects the points of $U$, and $\Pi(V)$ has objects the points of $V$; their common part is $\Pi(U\cap V)$.

Proposition: The groupoid $\Pi(X)$ is the pushout of $\Pi(U)$ and $\Pi(V)$ over $\Pi(U\cap V)$. In concrete terms, it is generated by paths in $U$ and paths in $V$, with their path-homotopy relations within each set and with a path in $U\cap V$ identified whether it is regarded as a path in $U$ or in $V$.

Proof. First, every morphism of $\Pi(X)$ is represented by a path $\alpha$. By path subdivision, $\alpha$ is a finite concatenation of paths lying in $U$ or $V$. This proves that the stated generators reach every morphism of $\Pi(X)$.

For the relations, take two concatenations of paths in the cover sets that represent the same morphism in $\Pi(X)$. Their concatenation, with one path reversed, is null-homotopic relative to its endpoints. Choose such a homotopy $H$ on the square. Subdivide the square into a grid whose closed cells each map into $U$ or $V$, as above. Each cell boundary is null-homotopic in its assigned set: its image lies there, and the cell contracts to a point inside the cell. Hence its boundary relation is already one of the path-homotopy relations in that set.

Assemble these cell relations across the grid. An edge shared by two cells with the same assigned set cancels with its reverse using the path inverse relation in that set. If the cells have different assigned sets, the shared edge lies in both $U$ and $V$, so its two readings are identified by the relation from $\Pi(U\cap V)$. After all interior edges cancel, the remaining relation is the boundary relation for the original homotopy. Therefore the two concatenations are equal using only the stated generators and relations. This proves there are no additional relations, and proves the groupoid pushout claim. $\square$

To pass from this groupoid statement to the theorem, use the path-connectedness assumptions. For each point $z\in X$, choose a path $t_z$ from $x_0$ to $z$: choose it in $U$ if $z\in U\setminus V$, in $V$ if $z\in V\setminus U$, and in $U\cap V$ if $z\in U\cap V$. These choices exist by path-connectedness. If $f$ is a path in $U$ from $z$ to $w$, then $t_z*f*t_w^{-1}$ is a loop in $U$ based at $x_0$; the corresponding assertion holds for paths in $V$. For a path in the intersection, both readings give the same based loop in the intersection. Thus the groupoid generators and relations, expressed using the paths $t_z$, become exactly the generators from $\pi_1(U,x_0)$, $\pi_1(V,x_0)$, and the identifications from $\pi_1(U\cap V,x_0)$. The groupoid pushout therefore gives the group pushout stated in the theorem. $\square$

Reading the Pushout and Avoiding a Common Error

Worked Example: Translating a Pushout into a Presentation

Suppose two infinite cyclic groups are generated by $a$ and $b$, and a third infinite cyclic group, generated by $c$, maps into them by $c\mapsto a^2$ and $c\mapsto b^3$. The pushout identifies the two images of every power $c^n$. In particular it imposes $a^2=b^3$; for every integer $n$, the resulting equality $(a^2)^n=(b^3)^n$ follows from this relation and its inverse.

The free product of the two cyclic groups has presentation $\langle a,b\mid\ \rangle$. Imposing the identification gives $$ \langle a,b\mid a^2=b^3\rangle. $$ There are no further relations from the common cyclic group: its elements are exactly the powers $c^n$, and each required identification is the $n$-th power of $a^2=b^3$. This is how the pushout in the theorem becomes a generators-and-relations description.

Common pitfall. A loop subdivided into paths in $U$ and $V$ cannot be treated as a product of elements of $\pi_1(U,x_0)$ and $\pi_1(V,x_0)$ without further work. A piece may start and end away from $x_0$, so it is not a based loop. The correction is to use the groupoid argument, or to choose paths $t_z$ from $x_0$ to each intermediate endpoint $z$ and convert a piece $f:z\to w$ into $t_z*f*t_w^{-1}$. When the transition point lies in $U\cap V$, the paths used to make this conversion can be chosen in the intersection. That is why the intersection’s path-connectedness is essential to this based-group formulation.

The proof separates the two jobs that the theorem must do. Path subdivision shows that local paths generate all global paths; homotopy subdivision shows that local path-homotopy relations generate all global relations. The shared edges of a grid account for the identifications on the overlap, and the group pushout records those identifications algebraically.

Check Your Understanding

Use the subdivision proof to check what each hypothesis or step contributes.

Why must the inverse images of $U$ and $V$ under a path be open for the path-subdivision argument?
In the homotopy-grid proof, why does a shared edge between cells assigned different cover sets lie in $U\cap V$?
What fails in the based-loop conversion if an intermediate endpoint cannot be joined to $x_0$ by a path in the relevant cover set?
For the cyclic-group maps $c\mapsto a^2$ and $c\mapsto b^3$, write the identification required for $c^{-2}$.
Explain why proving that every loop is a concatenation of local paths is not, by itself, enough to prove the Seifert–van Kampen theorem.

The Loop-Subdivision Proof of Van Kampen

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