The theoretical setting
Assume $ X_{-1}:=\emptyset, X_0 \subset X_1 \subset \dots \subset X\;$ is a CW-complex with pushouts
$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % \bigsqcup_{e \in \pi_0(X_{i}\setminus X_{i-1})} S^{i-1} & \ra{} & X_{i-1} \\ \da{\mathrm{inc}} & & \da{\mathrm{inc}}\\ \bigsqcup_{e \in \pi_0(X_{i}\setminus X_{i-1})} D^{i} & \ras{} & X_{i}\\ \end{array}\quad,\;i \in \mathbb{N}^\times \\ % $$
In a lecture I attended, we defined its (geometric) chain complex as follows:
For $i\in \mathbb{N}^\times$, choose some homeomorphism $\kappa_i$: $$ \begin{gather} D^i/S^{i-1} \xrightarrow[\kappa_i]{\cong} S^i \end{gather} $$
If $i \in \mathbb{N}^\times$, given the pushouts $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\la}[1]{\kern-1.5ex\xleftarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\las}[1]{\kern-1.5ex\xleftarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} %left diagram, upper row \bigsqcup_{e \in \pi_0(X_{i}\setminus X_{i-1})} S^{i-1} & \ra{\hspace{0.10cm} <\varphi_e>_e \hspace{0.1cm}} & X_{i-1} &% %right diagram, upper row \bigsqcup_{e \in \pi_0(X_{i+1}\setminus X_{i})} S^{i} & \ra{\hspace{0.10cm} <\varphi'_e>_e \hspace{0.1cm}} & X_{i} \\% %left diagram, middle row \da{\mathrm{inc}} & & \da{\mathrm{inc}} &% %right diagram, middle row \da{\mathrm{inc}} & & \da{\mathrm{inc}}\\% %left diagram, bottom row \bigsqcup_{e \in \pi_0(X_{i}\setminus X_{i-1})} D^{i} & \ras{\hspace{0.1cm}<\phi_e>_e\hspace{0.1cm}} & X_{i} &% %right diagram, bottom row \bigsqcup_{e \in \pi_0(X_{i+1}\setminus X_{i})} D^{i+1} & \ras{\hspace{0.1cm}<\phi'_e>_e\hspace{0.1cm}} & X_{i+1}\\ \end{array}\quad,\;i \in \mathbb{N}^\times \\ % $$
consider for given $ e \in \pi_0(X_{i+1}\setminus X_{i}),\,% e' \in \pi_0(X_{i}\setminus X_{i-1}) $ the morphism $f^{(i+1)}_{(e',e)}$ defined by going through the following (commutative) diagram from left to right: $$ \newcommand{\ra}[1]{\xrightarrow{\ \ #1\ \ }\phantom{}} \newcommand{\ras}[1]{\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}} \newcommand{\tra}[1]{\xtwoheadedrightarrow{\ \ #1\ \ }\phantom{}} \newcommand{\tras}[1]{\xtwoheadedrightarrow{\ \ \smash{#1}\ \ }\phantom{}} \newcommand{\la}[1]{\xleftarrow{\ \ #1\ \ }\phantom{}} \newcommand{\las}[1]{\xleftarrow{\ \ \smash{#1}\ \ }\phantom{} } \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\uda}[1]{\bigg\updownarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % S^i & \ra{\varphi_e} & X_{i} & \ra{\mathrm{can}} & X_{i}/X_{i-1} & \underset{\cong}{\la{\overline{<\phi'_{\varepsilon}>_{\varepsilon}}}} & \bigvee_{\varepsilon \in \pi_0(X_{i}\setminus X_{i-1})} D^i/S^{i-1} \\ & & \uda{=} & & \da{} & & \da{\mathrm{pr}_{e'}}\\ & & X_{i} & \ra{\mathrm{can}} & X_{i}/(X_{i}\setminus{e'}) & \ra{\phi'_{e'}} & D^i/S^{i-1} & \underset{\kappa_i}{\ra{\cong}} & S^i\\ \end{array}\quad,\;i \in \mathbb{N}^\times \\ % $$
Define for $i \in \mathbb{N}$ the cellular chain complexes $C^{\mathrm{cell,\, geo}}_{i+1}(X),\,C^{\mathrm{cell,\, geo}}_{i}(X)$ in the dimensions $i+1, i$ together with the boundary homomorphism: $$ \begin{align} \bigoplus_{e \in \pi_0(X_{i+1}\setminus X_{i})} \mathbb{Z} & \longrightarrow \bigoplus_{e' \in \pi_0(X_{i}\setminus X_{i-1})} \mathbb{Z} \\ \left(l_e\right)_e & \mapsto \left( \sum_{e \in \pi_0(X_{i+1}\setminus X_{i})} \deg\left(f^{i+1}_{e',\,e}\right) \cdot l_e \right)_{e' \in \pi_0(X_{i}\setminus X_{i-1})}\\ \end{align} $$
We have shown, that we then have an isomorphy of chain complexes $$ C^{\mathrm{cell,\, geo}}_\ast(X) \cong C^{\mathrm{cell}}_\ast(X):= H_\ast(X_\ast,\,X_{\ast-1}) $$ where the boundary homomorphisms for the right complex come from the exact homology sequences for the respective involved triples in the obvious way. I'm not sure if that makes a difference here, but $H_*$ denotes singular homology.
Now, with scalar multiplication as a group action, define the orbit spaces $ \mathbb{R}\mathrm{P}^j := (\mathbb{R}^{j+1}\setminus\{0\})/{\mathbb{R}^\times}\,,\;j\in \mathbb{N} $ and consider, for $n \in \mathbb{N}$ , the CW structure on $\mathbb{R}\mathrm{P}^{n+1}$ given by the pushouts
$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % S^{j} & \ra{\varphi^{j+1}\overset{\text{mod. homeomorphy}}= \mathrm{inc}} & \mathbb{R}\mathrm{P}^{j} \\ \da{\mathrm{inc}} & & \da{\mathrm{inc}}\\ D^{j+1} & \ras{\phi^{j+1}} & \mathbb{R}\mathrm{P}^{j+1} \\ & (x_0,\, \dots ,\, x_j) \mapsto \left[\left( x_0,\,\dots,\,x_i,\,\sqrt{1 - \Sigma_i x_i^2} \right)\right] \end{array}\quad,\; 0 \leq j \leq n \\ % $$
where the attaching map comes from restriction. We can homoemorphically identify $\mathbb{R}\mathrm{P}^j = \left( \mathbb{R}^{j+1}\times \{0\} \setminus \{0\} \right)/\mathbb{R}^\times = can(\mathbb{R}^{j+1}\times \{0\} \setminus \{0\}) \subset \mathbb{R}\mathrm{P}^{j+1}$) of the lower map and with these identifications, we have a CW filtration
$$ X_{-1}:=\emptyset \; \subset \; X_0=\mathbb{R}\mathrm{P}^0 \; \subset \; X_1=\mathbb{R}\mathrm{P}^1 \; \subset \; \dots \; \subset \; \mathbb{R}\mathrm{P}^{n+1} = X_{n+1} = X $$
giving the chain complex with a space isomorphic to $\mathbb{Z}$ in each dimension $0 \leq j \leq n+1$ and to compute the homology of $X$, we have to compute the homology of this chain complex; this means we need to compute the boundary homomorphisms.
The argument
In the lecture notes, which to my dismay are very much following Hatcher's AT (not my style, I like it more formally), the argument is that we have a diagram
$$ \newcommand{\ra}[1]{\xrightarrow{\ \ #1\ \ }\phantom{}} \newcommand{\ras}[1]{\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}} \newcommand{\tra}[1]{\xtwoheadedrightarrow{\ \ #1\ \ }\phantom{}} \newcommand{\tras}[1]{\xtwoheadedrightarrow{\ \ \smash{#1}\ \ }\phantom{}} \newcommand{\la}[1]{\xleftarrow{\ \ #1\ \ }\phantom{}} \newcommand{\las}[1]{\xleftarrow{\ \ \smash{#1}\ \ }\phantom{} } \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\uda}[1]{\bigg\updownarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % S^i & \ra{\mathrm{can}} & S^i/S^{i-1} & \underset{\cong}{% \la{\overline{\mathrm{inc}}\vee\overline{\mathrm{inc}}}% } & D_+^i/S^{i-1} \vee D_-^i/S^{i-1} \\ \da{\mathrm{can}' \circ \mathrm{inc}} & & \da{\overline{\mathrm{can}_i\circ\mathrm{inc}}} & & \da{\mathrm{id} \vee \bar\tau} & & \\ \mathbb{R}\mathrm{P}^{i} & \ra{\mathrm{can}_i} & \mathbb{R}\mathrm{P}^{i}/\mathbb{R}\mathrm{P}^{i-1} & \underset\cong{\la{\mathrm{can}}} & D_+^i/S^{i-1} \cong D^i/S^{i-1} & \underset{\kappa_i}{\ra{\cong}} & S^i\\ \end{array}\quad,\;i \in \mathbb{N}^\times \\ % $$
where
- as in the other diagrams, $\mathrm{can}, \mathrm{inc}, \mathrm{pr}$ stand for the usual inclusions/canonical projections/projections into one of the wedge factors (modulo adding some index subscript to make the maps distinguishable) and overlines indicate maps 'that have descended'
- $D_\pm^i \subset S^i$ stands for the upper/lower hemisphere which we can both homeomorphically identify with $D^i$ by 'downward' projection onto the obvious hyperplane
- $\tau$ is the antipodal map
In the lecture notes, we are using local degrees to calculate the degree (using 'the' degree formula by focusing on small open neighborhoods of the north/southpole $n$ respectively $s$ of $S^i$ and suppressing the cell index in the notation of $f^{i+1}$ (see above for the definition of $f^{i+1}$),)
$$
\mathrm{deg}(f^{i+1}) = \mathrm{deg}(f^{i+1}, n) + \mathrm{deg}(f^{i+1}, s) = 1 + (-1)^{i+1}
$$
The problem I am facing
I understand that due to the given diagram, $f$ intuitively acts somewhat like the identity around $n$, and somewhat like the antipodal map around $s$. But this isn't a convincing argument for me.
How can one technically/more formally prove this?
My approach so far was actually not using local degrees, as I feel they only complicate matters here, but it's quite similar. I tried using the definition of singular homology as classes of chain simplices. Applying the homology functor:
$$ \newcommand{\ra}[1]{\xrightarrow{\ \ #1\ \ }\phantom{}} \newcommand{\ras}[1]{\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}} \newcommand{\tra}[1]{\xtwoheadedrightarrow{\ \ #1\ \ }\phantom{}} \newcommand{\tras}[1]{\xtwoheadedrightarrow{\ \ \smash{#1}\ \ }\phantom{}} \newcommand{\la}[1]{\xleftarrow{\ \ #1\ \ }\phantom{}} \newcommand{\las}[1]{\xleftarrow{\ \ \smash{#1}\ \ }\phantom{} } \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \newcommand{\uda}[1]{\bigg\updownarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % H_i(S^i) & \ra{\mathrm{can}*} & H_i(S^i/S^{i-1}) & \underset{\cong}{% \ra{(\overline{\mathrm{pr_+}}_*,\,\overline{\mathrm{pr_-}}_*)}% } & H_i(D_+^i/S^{i-1}) \oplus H_i(D_-^i/S^{i-1}) \\ & & & & \da{<1 ,\, \bar\tau_*>} & & \\ & & & & H_i(D_+^i/S^{i-1}) \cong H_i(D^i/S^{i-1}) & \underset{(\kappa_i)_*}{\ra{\cong}} & H_i(S^i)\\ \end{array}\quad,\;i \in \mathbb{N}^\times \\ % $$
where
- the brackets indicate the homomorphisms owing to the property of the direct sum of modules
- the parentheses indicate homomorphisms due to the direct product of modules
- $pr_\pm$ is the projection $S^i \rightarrow D_\pm^i/S^{i-1}$
So I can look at what the induced map does on the class of a singular simplex $\sigma$, namely $$ [\sigma] \mapsto \mathrm{can}_\ast([\sigma)]) \mapsto \left(% (\overline{pr_+}_\ast\circ\mathrm{can}_\ast)[\sigma],\;% (\overline{pr_-}_\ast\circ\mathrm{can}_\ast)[\sigma] \right) \mapsto (\overline{pr_+}_\ast\circ\mathrm{can}_\ast)[\sigma] + % (\overline{pr_-}_\ast\circ\mathrm{can}_\ast)[\sigma] \\ = (\overline{pr_+}_\ast\circ\mathrm{can}_\ast)[\sigma] + % (\overline{pr_+}_\ast\circ\bar\tau_\ast\circ\mathrm{can}_\ast)[\sigma] \\ = [\overline{pr_+} \circ \mathrm{can} \circ \sigma] + % [\overline{pr_+} \circ \bar\tau \circ \mathrm{can} \circ \sigma] $$
Now, I am not sure how to proceed.
- I do know, that the antipodal map on $\mathbb{R}^{i+1}$ has degree $(-1)^{i+1}$; but $\tau$ is not that antipodal map right here; we are looking at the quotient of an antipodal map. So how do prove that $\bar\tau_* = (-1)^{i+1} ( \cdot )$?
- Since I need to show this: What is a (good) way to show that $[\overline{pr_+} \circ \mathrm{can} \circ \sigma]$ is a generator, if $[\sigma]$ is?
I'm a little tired, but I suppose if I can answer these questions, then, since we are dealing with linear maps, I'm done.
I'm just having a hard time seeing how the maps that induce the homological morphisms give these results...
Please give a technical/more concrete answer, not a 'Hatcheresque' one. I don't need intuitive images that much, I want a formal argument. If there is yet another way, I'm also open to a formal argument making use of the local degree formula, just please be explicit.