Talk:Reducible Poincaré Complexes (Ex)

1. Let's quickly prove the mentioned theorem of Wall. Assume that $X$$<wikitex>; 1. Let's quickly prove the mentioned theorem of Wall. Assume that $X$ is connected. The collection of top-dimensional $n$-cells $(e^{n}_{i},\phi_{i})_{i\in\{1,\ldots,k\}}$ is such that $\bigcup_{i\in\{1,\ldots,k\}} \phi_{i}(\mathbb{S}^n)$ is connected - otherwise any two connected components would give two independent classes in $H_{n}(X)$. Assume that $X$ is pointed and that all attaching maps are pointed as well. We can take $X^{\bullet}$ to be the $(n-1)$-skeleton of $X$ and the unique $n$-cell to have the attaching map $$\phi\colon \mathbb{S}^{n}\xrightarrow{}\bigvee_{i\in\{1,\ldots,k\}}\mathbb{S}^{n}\xrightarrow{\phi_1\vee\ldots\vee\phi_k}X^{\bullet}$$ begginning with the pinching map. </wikitex>X$ is connected. The collection of top-dimensional $n$$n$-cells $(e^{n}_{i},\phi_{i})_{i\in\{1,\ldots,k\}}$$(e^{n}_{i},\phi_{i})_{i\in\{1,\ldots,k\}}$ is such that $\bigcup_{i\in\{1,\ldots,k\}} \phi_{i}(\mathbb{S}^n)$$\bigcup_{i\in\{1,\ldots,k\}} \phi_{i}(\mathbb{S}^n)$ is connected - otherwise any two connected components would give two independent classes in $H_{n}(X)$$H_{n}(X)$. Assume that $X$$X$ is pointed and that all attaching maps are pointed as well. We can take $X^{\bullet}$$X^{\bullet}$ to be the $(n-1)$$(n-1)$-skeleton of $X$$X$ and the unique $n$$n$-cell to have the attaching map

$\displaystyle \phi\colon \mathbb{S}^{n}\xrightarrow{}\bigvee_{i\in\{1,\ldots,k\}}\mathbb{S}^{n}\xrightarrow{\phi_1\vee\ldots\vee\phi_k}X^{\bullet}$ $$\displaystyle \phi\colon \mathbb{S}^{n}\xrightarrow{}\bigvee_{i\in\{1,\ldots,k\}}\mathbb{S}^{n}\xrightarrow{\phi_1\vee\ldots\vee\phi_k}X^{\bullet}$$

begginning with the pinching map.