Ytukyg

Suppose the topological space $X$ is the union of two open pathwise connected subspaces $U_1$ and $U_2$ whose intersection is also pathwise connected. Let $xin U_1cap U_2$ and let $f_k:pi_1(U_k,x)to G$ ($k=1,2$) be a homomorphism, where $G$ is a group. Assume $f_{U_1}circ (i_{U_1})_ast=f_{U_2}circ (i_{U_2})_ast$, where the $i_{U_k}$ are the inclusions. Consider a loop $gamma:Ito X$ based at $x$; assume $I$ is divided into subintervals $I_j=[x_j,x_{j+1}]$ ($jin Lambda$) so that the image of $gammarestriction_{I_j}$ lies entirely within $U_1$ or $U_2$. Let $beta_i: Ito U_1cap U_2$ be a path with $beta_i(0)=x,beta_i(1)=gamma(x_i)$. Then the concatenation $Gamma_i=beta_{i-1} cdot gammarestriction_{I_{i+1}}cdot beta_i^{-1}$ is a path in $U_k$ for some $kin{1,2}$. And we can consider the product of $f_{U_i}([Gamma_i])$ over all $iin Lambda$. (If $Gamma_i$ lies in $U_i$, then the image under $f_{U_i}$ (of $[Gamma_i]$) is taken.)

I'm trying to understand why the value of the product is independent of the choice of the partition of $I$ into the $I_j$ as well as of the choice of $beta_i$. I have no idea about the former, and the latter doesn't even look plausible for me. If $delta_i$ is a path which enjoys the properties of $beta_i$, then I guess I have to show that $delta_i$ is fixed-end-point homotopic to $beta_i$ (or what else can I show?), but the intersection needn't be simply connected.. So the independence of both choices looks quite mysterious for me.

We need some more precision.

1) Let $I$ be divided into $n$ subintervals $I_j = [x_{j-1},x_j]$, $j = 1,dots,n$, such that $gamma(I_j)$ is contained in $U_1$ or $U_2$.

2) For $j = 1,dots,n-1$ let $beta_j : I to U_1 cap U_2$ be a path such that $beta_j(0) = x, beta_j(1) = gamma(x_j)$. For $j = 0, n$ let $beta_j$ be the constant path $beta_j(t) equiv x$.

3) For each $j = 1dots,n$ choose $k(j) in { 1,2 }$ such that $gamma(I_j) subset U_{k(j)}$. Note that $gamma(I_j) subset U_1 cap U_2$ is possible. For $j = 1,dots,n$ define a closed path in $U_{k(j)}$ by
$$Gamma_j = Gamma_j(beta_{j-1},beta_j) = beta_{j-1} cdot gamma mid_{I_j} cdot beta^{-1}_j .$$
Then define
$$f(gamma) = f_{k(1)}([Gamma_1]) dots f_{k(n)}([Gamma_n]) in G.$$

The claim is that $f(gamma)$ does not depend on the above choices.

a) Independence of the choice of $beta_j$.

Recall that a choice was made only for $j = 1,dots,n-1$. Thus it suffices to show that for $j = 1,dots,n-1$
$$f_{k(j)}([Gamma_j(beta_{j-1},beta_j)]) f_{k(j+1)}([Gamma_{j+1}(beta_j,beta_{j+1})]) = f_{k(j)}([Gamma_j(beta_{j-1},beta'_j)]) f_{k(j+1)}([Gamma_{j+1}(beta'_j,beta_{j+1})]).$$
The path $delta_j = beta'_j cdot beta^{-1}_j$ is a closed path in $U_1 cap U_2$ which begins and ends at $x$. Let $i_k : U_1 cap U_2 to U_k$ denote inclusion. We write
$$phi = f_1 circ (i_1)_* = f_2 circ (i_2)_* .$$
We have
$$[Gamma_j(beta_{j-1},beta'_j)] cdot [i_{k(j)}delta_j] = [Gamma_j(beta_{j-1},beta_j)]$$
and therefore
$$f_{k(j)}([Gamma_j(beta_{j-1},beta_j)]) = f_{k(j)}([Gamma_j(beta_{j-1},beta'_j)]) f_{k(j)}([i_{k(j)}delta_j]) = f_{k(j)}([Gamma_j(beta_{j-1},beta'_j)]) f_{k(j)}(i_{k(j)})_*([delta_j])$$
$$= f_{k(j)}([Gamma_j(beta_{j-1},beta'_j)]) phi([delta_j]) .$$
Similarly we see that
$$ f_{k(j+1)}([Gamma_{j+1}(beta_j,beta_{j+1})]) = phi([delta_j])^{-1} f_{k(j+1)}([Gamma_{j+1}(beta'_j,beta_{j+1})])$$
which completes the proof of a).

b) Independence of the choice of $k(j)$.

A choice is only possible when $gamma(I_j) subset U_1 cap U_2$. In that case we use $f_1 circ (i_1)_* = f_2 circ (i_2)_*$.

c) Independence of the choice of the partition of $I$.

Let us first consider a refinement of ${ I_j }$ by inserting one additional partition point between $x_j$ and $x_{j+1}$ for some $j$. By considerations similar as in a) we see that this does not change $f(gamma)$. Procceeding inductively we see that the same is true for any refinement of ${ I_j }$. Now any two partitions have a common refinement which proves c).

Note that the proof has to be continued by showing that $f(gamma)$ depends only on $[gamma]$. But that is another story.