# Product of paths is independent of choices

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 $$x\in U_1\cap U_2$$ 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 $$\gamma:I\to X$$ based at $$x$$; assume $$I$$ is divided into subintervals $$I_j=[x_j,x_{j+1}]$$ ($$j\in \Lambda$$) so that the image of $$\gamma\restriction_{I_j}$$ lies entirely within $$U_1$$ or $$U_2$$. Let $$\beta_i: I\to U_1\cap U_2$$ be a path with $$\beta_i(0)=x,\beta_i(1)=\gamma(x_i)$$. Then the concatenation $$\Gamma_i=\beta_{i-1} \cdot \gamma\restriction_{I_{i+1}}\cdot \beta_i^{-1}$$ is a path in $$U_k$$ for some $$k\in\{1,2\}$$. And we can consider the product of $$f_{U_i}([\Gamma_i])$$ over all $$i\in \Lambda$$. (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.

• The point about the $\beta_i$ is that in $\Gamma_i\cdot \Gamma_{i+1}$, they cancel. Actually, that same argument (with a "common refinement") is also what shows independence of the partition – Hagen von Eitzen Nov 26 '18 at 4:51
• @HagenvonEitzen But to collect $\Gamma_i$ and $\Gamma_{i+1}$ in one place, one has to apply the homomorphism property (I believe); however it may be the case that the product is of the form $f_{U_1}(..)f_{U_2}(..)f_{U_1}(..)...$, so the homomorphism property cannot be applied. (Also, the group isn't commutative.) And even if it did, I don't quite see how this shows independence. – user531587 Nov 26 '18 at 5:05

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 $$j = 1\dots,n$$ 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.

• Thank you! I like the style of your proof very much - you are very precise. There is one thing I'm not sure I understand though, namely why is it sufficient to show what you are claiming (in the proof of (a))? Also, as a bonus question, the discussion in the comments above alludes to the fact that the map that assigns to $\gamma$ the corresponding product behaves nicely under products of paths (has the "homomorphism property"). Is that indeed the case? (Please let me know if this deserves a separate question, which I will then ask.) – user531587 Nov 27 '18 at 3:29
• @user531587 Concerning a): We know that $f_{k(j)}([\Gamma_j(\beta_{j-1},\beta_j)]) f_{k(j+1)}([\Gamma_{j+1}(\beta_j,\beta_{j+1})]) = f_{k(j)}([\Gamma_j(\beta_{j-1},\beta'_j)]) f_{k(j+1)}([\Gamma_{j+1}(\beta'_j,\beta_{j+1})])$. This means that in $f_{k(1)}([\Gamma_1]) \dots f_{k(n)}([\Gamma_n])$ which is based on $\beta_1,\dots,\beta_{n-1}$ (recall that $\beta_0,\beta_n$ are constant!) we can replace any $\beta_j$ by $\beta'_j$ without changing the product. – Paul Frost Nov 27 '18 at 9:53
• Concerning your bonus question: Given closed paths $\gamma_1, \gamma_2$, the path $\gamma =\gamma_1 \cdot \gamma_2$ is defined by $\gamma(t) =\gamma_1(2t)$ for $t \le 1/2$ and $\gamma(t) =\gamma_2(2t-1)$ for $t \ge 1/2$. Now choose a partition of $I$ such that $1/2$ is a partition point. Then you will easily see that $f(\gamma) = f(\gamma_1) f(\gamma_2)$. – Paul Frost Nov 27 '18 at 9:53