van Kampen’s Theorem

We present a variant of Hatcher’s proof of van Kampen’s Theorem, for the simpler case of just two open sets.

Theorem 1 Let X be a space with basepoint x_{0}. Let A_{1} and A_{2} be open subspaces that contain x_{0} and satisfy X = A_{1} ∪ A_{2}. Assume that A_{1}, A_{2} and A_{1} ∩ A_{2} (and hence X) are all path-connected. Then the commutative square of homomorphisms induced by inclusions in the diagram

\xymatrix{ &  & G \\ π_{1}(A_{1}) \ar[r]_{j_{1∗}} \ar[urr]^{ϕ_{1}} & π_{1}(X) \ar@{-->}[ur]^{ϕ} \\ π_{1}(A_{1} ∩ A_{2}) \ar[u]^{i_{1∗}} \ar[r]^{i_{2∗}} & π_{1}(A_{2}) \ar[u]^{j_{2∗}} \ar[uur]_{ϕ_{2}}}  (2)

is a pushout square of groups: given any group G and homomorphisms ϕ_{1}: π_{1}(A_{1}) → G and ϕ_{2}: π_{1}(A_{2}) → G such that ϕ_{1} ^{∘} i_{1∗} = ϕ_{2} ^{∘} i_{2∗}, there is a unique homomorphism ϕ: π_{1}(X) → G that makes the diagram commute, i. e. ϕ ^{∘} j_{1∗} = ϕ_{1} and ϕ ^{∘} j_{2∗} = ϕ_{2}.

The proof will involve subdivisions of paths. Take any path f: I → X in X and let 0 = s_{0} < s_{1} < s_{2} < … < s_{m−1} < s_{m} = 1 be any subdivision of I. Write f_{i} for the obvious path from f(s_{i−1}) to f(s_{i}), namely f_{i}(s) = f((1−s)s_{i−1} + ss_{i}).

Lemma 3 With the above notation, there is a path homotopy f ≃ f_{1} · f_{2} · . . . · f_{m}.

Remark The convention here is that parentheses are to be inserted anywhere in f_{1} · f_{2} · . . . · f_{m} to make it defined; since all choices yield path-homotopic results, the specific choice is irrelevant.

Proof Given a, b ∈ I, denote by λ_{a,b}: I → I the straightline path in I from a to b, i. e. λ_{a,b}(s) = (1−s)a + sb. Then f_{i} = f ^{∘} λ_{s_{i−1},s_{i}} and

f_{1} · f_{2} · . . . · f_{m} = (f^{∘}λ_{s_{0},s_{1}}) · (f^{∘}λ_{s_{1},s_{2}}) · . . . · (f^{∘}λ_{s_{m−1},s_{m}})
= f ^{∘} (λ_{s_{0},s_{1}} · λ_{s_{1},s_{2}} · . . . · λ_{s_{m−1},s_{m}})
≃ f ^{∘} id_{I} = f,

since λ_{s_{0},s_{1}} · λ_{s_{1},s_{2}} · . . . · λ_{s_{m−1},s_{m}} ≃ id_{I}, I being convex.

Proof of Theorem Because we have to deal with four different spaces, we write [f]_{Y} for the path-homotopy class in Y of a loop in Y. (The basepoint for all loops will be x_{0} and is suppressed from the notation.)

We have to construct ϕ(f) for each loop f in X at x_{0}.

The recipe Given f, the open sets f^{−1}(A_{1}) and f^{−1}(A_{2}) cover the compact metric space I. By the Lebesgue covering lemma, there is a subdivision 0 = s_{0} < s_{1} < s_{2} < … < s_{m} = 1 of I such that for each i, f([s_{i−1}, s_{i}]) ⊂ A_{α(i)}, where α(i) = 1 or 2.

[figure: …; s_{0} = 0; f_{1}; s_{1}; f_{2}; s_{2}; s_{m−1}; f_{m}; s_{m} = 1]

We note that f(s_{i}) ∈ A_{α(i)} ∩ A_{α(i+1)}. Since this space is path-connected, we can choose, for each i (0 < i < n), a path g_{i} in A_{α(i)} ∩A_{α(i+1)} from x_{0} to f(s_{i}); we also put g_{0} = g_{m} = c, the constant loop at x_{0}. For each i, we have the loop ω_{i} = g_{i−1} · f_{i} · \overline{g}_{i} in A_{α(i)}. We put z_{i} = ϕ_{α(i)}([ω_{i}]_{A_{α(i)}}) and set

ϕ(f) = z_{1}z_{2} … z_{m} in G.  (4)

Commutativity Given a loop f in A_{β}, we take m = 1 and don’t subdivide at all, 0 = s_{0} < s_{1} = 1. Then ω_{1} ≃ f and (4) reduces to ϕ(f) = ϕ_{β}([f]_{A_{β}}), which will imply that ϕ(j_{β∗}([f]_{A_{β}})) = ϕ(f) = ϕ_{β}([f]_{A_{β}}), as required.

Uniqueness of ϕ[f] By Lemma 3,

[ω_{1}]_{X}[ω_{2}]_{X} … [ω_{m}]_{X} = [c · f_{1} · \overline{g}_{1} · g_{1} · f_{2} · \overline{g}_{2} · . . . · g_{m−1} · f_{m} · c]_{X} = [f]_{X}

in π_{1}(X). Since ϕ is to be a homomorphism, we must use the formula (4), with

z_{i} = ϕ([ω_{i}]_{X}) = ϕ(j_{α(i)∗}([ω_{i}]_{A_{α(i)}})) = ϕ_{α(i)}([ω_{i}]_{A_{α(i)}}).

Homomorphism To verify that ϕ is a homomorphism on π_{1}(X), we have to show that, given two loops f and f′, ϕ(f · f′) = ϕ(f)ϕ(f′). Choose subdivisions, α(i) and g_{i} as above for each of f and f′, to get elements z_{i} and z′_{j} of G. If we combine these subdivisions to get a subdivision for f · f′, we find

ϕ(f · f′) = z_{1}z_{2} … z_{m}z′_{1}z′_{2} … z′_{m′} = ϕ(f)ϕ(f′).

To prove that ϕ[f] is well defined on π_{1}(X) (which will complete the proof), we have to show it is independent of:

(i) The choice of α(i) for each i; (ii) The choice of g_{i} for each i; (iii) The choice of subdivision of I; (iv) The choice of f in the homotopy class [f].

ϕ(f) is independent of the choice of α(i). There is no choice unless f_{i} lies in A_{1} ∩A_{2}. In this case, ω_{i} also lies in A_{1} ∩ A_{2} and z_{i} is well defined, because

ϕ_{1}([ω_{i}]_{A_{1}}) = ϕ_{1}(i_{1∗}([ω_{i}]_{A_{1}∩A_{2}})) = ϕ_{2}(i_{2∗}([ω_{i}]_{A_{1}∩A_{2}})) = ϕ_{2}([ω_{i}]_{A_{2}}).

ϕ(f) is independent of g_{i}. Suppose we choose a different path g′_{i} from x_{0} to f(s_{i}) in A_{β} ∩ A_{γ}, where α(i) = β and α(i+1) = γ. We then have the loop k = g_{i} · \overline{g}′_{i}, so that \overline{k} = g′_{i} · \overline{g}_{i}. We replace ω_{i} by ω′_{i} = g_{i−1} · f_{i} · \overline{g}′_{i} ≃ ω_{i} · k and ω_{i+1} by ω′_{i+1} = g′_{i} · f_{i+1} · \overline{g}_{i+1} ≃ \overline{k} · ω_{i+1}. The other loops ω_{j} are unchanged. Then

z′_{i}z′_{i+1} = z_{i}ϕ_{β}([k]_{A_{β}})ϕ_{γ}([\overline{k}]_{A_{γ}})z_{i+1} in G.

Since [\overline{k}] = [k]^{−1}, this reduces to z_{i}z_{i+1} if β = γ. Otherwise, we use

ϕ_{1}([k]_{A_{1}}) = ϕ_{1}(i_{1∗}([k]_{A_{1}∩A_{2}})) = ϕ_{2}(i_{2∗}([k]_{A_{1}∩A_{2}})) = ϕ_{2}([k]_{A_{2}}).

ϕ(f) is independent of the subdivision. To compare any two subdivisions, it is enough to consider the effect of inserting one extra point s_{+} between s_{i−1} and s_{i}. Suppose f_{i} maps into A_{β}; then we break up f_{i} ≃ f′_{i} · f′′_{i} as in Lemma 3, where f′_{i} and f′′_{i} also map into A_{β}. We choose a path g_{+} in A_{β} from x_{0} to f(s_{+}); then in (4), we replace z_{i} by

z′_{i}z′′_{i} = ϕ_{β}([g_{i−1} · f′_{i} · \overline{g}_{+}]_{A_{β}})ϕ_{β}([g_{+} · f′′_{i} · \overline{g}_{i}]_{A_{β}})
= ϕ_{β}([g_{i−1} · f′_{i} · \overline{g}_{+} · g_{+} · f′′_{i} · \overline{g}_{i}]_{A_{β}})
= ϕ_{β}([ω_{i}]_{A_{β}}) = z_{i},

which does not change ϕ(f).

Homotopy To make ϕ well defined on π_{1}(X), we have to show that for any loop homotopy f_{t}, ϕ(f_{0}) = ϕ(f_{1}). The homotopy provides a map F: I × I → X. The open sets F^{−1}(A_{1}) and F^{−1}(A_{2}) cover the compact metric space I × I. Again by the Lebesgue covering lemma, we can subdivide the path variable s ∈ I by

0 = s_{0} < s_{1} < … < s_{m} = 1  (5)

and the homotopy variable t ∈ I by 0 = t_{0} < t_{1} < … < t_{n} = 1 to make each rectangle R_{i,j} = [s_{i−1}, s_{i}] × [t_{j−1}, t_{j}] map into A_{1} or into A_{2}; we have only to make them small enough. It is enough to show that ϕ(f_{t_{j−1}}) = ϕ(f_{t_{j}}) for each j.

Suppose F (R_{i,j}) ⊂ A_{β}; we use α(i) = β and the same subdivision (5) for both f_{t_{j−1}} and f_{t_{j}}, as pictured.

[figure: …; …; t_{j}; u′_{i}; v_{i−1}; R_{i,j}; v_{i}; …; …; t_{j−1}; u_{i}; s_{i−1}; s_{i}]

As in Lemma 3, we subdivide f_{t_{j−1}} ≃ u_{1} · u_{2} · . . . · u_{m} and f_{t_{j}} ≃ u′_{1} · u′_{2} · . . . · u′_{m}. Let v_{i} be the obvious path from F (s_{i}, t_{j−1}) to F (s_{i}, t_{j}). As usual, we choose paths g_{i} from x_{0} to f_{t_{j−1}}(s_{i}) for f_{t_{j−1}}; but for f_{t_{j}}, we use the paths g′_{i} = g_{i} · v_{i}. Then for (4), ω_{i} = g_{i−1} · u_{i} · \overline{g}_{i}, while ω′_{i} = g_{i−1} · v_{i−1} · u′_{i} · \overline{v}_{i} · \overline{g}_{i}. Because R_{i,j} is convex and maps into A_{β}, u_{i} ≃ v_{i−1} · u′_{i} · \overline{v}_{i} in A_{β}, which implies ω′_{i} ≃ ω_{i} and z′_{i} = z_{i}.

Extensions 1. The sets A_{1} and A_{2} do not need to be open; it is enough to have Int A_{1} ∪ Int A_{2} = X. To prove existence of a suitable subdivision of I or I × I, we start from the open covering of X by Int A_{1} and Int A_{2}. The rest of the proof is unchanged.

2. Instead of two open sets A_{1} and A_{2}, we can have many, even infinitely many. Beyond the obvious changes to the statement and proof, one extra condition is needed:

A_{α} ∩ A_{β} ∩ A_{γ} is path-connected for all α, β, γ.

Hatcher (p. 44) gives an example to show the necessity. See if you can find where in this proof it is needed.