Example of homomorphism $\underset{\alpha \in A}{\ast} \pi_1(U_\alpha) \to \pi_1(X)$, where $U_\alpha \ni x_0$ are an open cover of $(X,x_0)$.

Question

I’m reading a passage leading up to the Van Kampen’s theorem, and I have trouble understanding the settings. Here’s the materials.

Let us consider a pointed space $\left(X, x_{0}\right)$ with an open cover $\left\{U_{\alpha}: \alpha \in A\right\}$ by path-connected open sets $U_{\alpha} \ni x_{0}$ containing the basepoint. We fix the basepoint $x_{0}$ throughout, using it for $X$ and for each $U_{\alpha},$ as well as for their intersections; we omit it from the notation for fundamental groups. The inclusions $U_{\alpha} \hookrightarrow X$ induce homomorphisms $j_{\alpha}: \pi_{1}\left(U_{\alpha}\right) \rightarrow \pi_{1}(X) .$ By the universal property of the free product, these combine to give a homomorphism $$\varphi: F:=\underset{\alpha \in A}{*} \pi_{1}\left(U_{\alpha}\right) \rightarrow \pi_{1}(X)$$ Consider now, the inclusions $U_{\alpha} \cap U_{\beta} \hookrightarrow U_{\alpha}($ for $\alpha, \beta \in A)$. These induce homomorphisms $$ i_{\alpha \beta}: \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right) \rightarrow \pi_{1}\left(U_{\alpha}\right) $$

The inclusions of $U_{\alpha} \cap U_{\beta}$ into $X$ via $U_{\alpha}$ or $U_{\beta}$ commute:

By functoriality, the same is true of the induced homomorphisms:

That is, for all $\alpha, \beta \in A$ we have $$ j_{\alpha} \circ i_{\alpha \beta}=j_{\beta} \circ i_{\beta \alpha}: \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right) \rightarrow \pi_{1}(X) $$ This implies that for any loop $\sigma \in \pi_{1}\left(U_{\alpha} \cap U_{\beta}\right)$ we have $$ \varphi\left(i_{\alpha \beta}(\sigma)\right)=\varphi\left(i_{\beta \alpha}(\sigma)\right) \in \pi_{1}(X) $$ Thus the two-letter word $i_{\alpha \beta}(\sigma) i_{\beta \alpha}(\sigma)^{-1}=i_{\alpha \beta}(\sigma) i_{\beta \alpha}\left(\sigma^{-1}\right)$ in the free group $F= \underset{\alpha}{\ast} \pi_{1}\left(U_{\alpha}\right)$ is in the kernel of $\varphi$.

My questions are as follows:

1. Just to be sure, are the maps $i$ and $j$ just the inclusions of homotopy classes of loops from one fundamental group to another?
2. What does the map $\varphi$ do exactly? I have trouble understanding how the maps $j_\alpha$ and $j_\beta$ “combine” to give the map $\varphi$.
3. Perhaps an example would make the idea clearer. Please use the following setting if it makes sense: say $U_\alpha$ and $U_\beta$ are both homeomorphic to $S^1$, and $a$ and $b$ are the generators of their respective fundamental groups. A word in $F$ could be something like $a^3b^{-5}$. What are the remaining components of the second diagram above, and the corresponding image of $\varphi$? If this doesn’t make sense, please let me know why and offer another example if possible.

Answer 1

1. Yes

2. $\phi$ takes a free product (or word) $\Pi_\alpha \ \gamma_\alpha$ of loops $\gamma_\alpha \in \pi_1(U_\alpha)$ to the product $\Pi_\alpha \ j_\alpha(\gamma_\alpha) \in \pi_1(X)$ where the product in $\pi_1(X)$ is concatenation of loops. It's really only the maps $j$ which combine to give $\phi$.

3. This question doesn't really make sense, it's impossible to say what the space $X$ is from this situation. If $X = U_0 = U_1 = S^1$ then we have that $\phi(a^3b^{-5}) = a^{3-5} = a^{-2}$ assuming $a = b$.

More clearly, $\phi: \mathbb Z* \mathbb Z \rightarrow \mathbb Z$ in this situation is the identity on each of the copies of $\mathbb Z$ so takes a word of integers to its product (really sum) in $\mathbb Z$.

If you don't understand something I would be happy to elaborate.