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A while ago, I'd asked a related question here, which went unanswered probably because it was too broad a question.

Suppose that we have chain complex for a CW comples $X$: $\cdots\to C_{n+1}\to C_n\to C_{n-1}\to\cdots$ and let $d_j: C_j\to C_{j-1}$ be the boundary maps. Here $C_n=\mathbb Z^a, a=$ number of $n$ cells in $X$. $d_i$'s are computed by cellular boundary formula. Let me state my understanding of it: $d_j(e_\alpha^j)= \sum d_{\alpha\beta} e_\beta^{j-1},$ where $e_\alpha^j=j$ cell and $d_{\alpha\beta}=$ degree of the composition $(S^{n-1}\to X^{n-1}\to S^{n-1}),$ where the first map is the attaching map of the $n-$cell $e_\alpha^n$ and the second map is the quotient map collapsing $X^{n-1}-e_\beta^{n-1}$ to a point. I don't quite understand this second map. This is because on page $141$, Hatcher defines $d_{\alpha\beta}$ as degree of the composition $q_\beta qq_\alpha$, where

$a) q: X^{n-1}\to X^{n-1}/X^{n-2}$ is the quotient map.

$b)q_\beta: X^{n-1}/X^{n-2}\to S_\beta^{n-1}$ is the map collapsing the complement of $e_\beta^{n-1}$ to a point. I don't understand what this map does to rest of the space.

$c) \phi_\alpha: S^{n-1}\to X^{n-1}$ is the attaching map.

I don't understand how this definition of $d_{\alpha \beta}$ matches with the earlier definition.

For now, let's assume that somehow they are same and try to compute the homology groups of $RP^n$. I know that as a cell complex it has one $k-$ cell for $k\le n$. So I get a chain complex: $\cdots \to 0\to C_n\to C_{n-1}\to $, where $C_k=\mathbb Z$ for all $k\le n$. Now the task is to compute $d_k: C_k\to C_{k-1}$.

Here is what Hatcher does:

the attaching map for $e^k$ is the $2$ sheeted covering projection $ϕ: S^{k−1}→RP^{k−1}$. To compute the boundary map $d_k$ we compute the degree of the composition $S^{k−1}\overbrace{\to}^{\phi} RP^{k−1}\overbrace{\to}^q RP^{k−1}/RP^{k−2} = S^{k−1}$ , with $q$ the quotient map. The map $qϕ$ restricts to a homeomorphism from each component of $S^{k−1} − S^{k−2}$ onto $RP^{k−1} −RP^{k−2}$ , and these two homeomorphisms are obtained from each other by precomposing with the antipodal map of $S^{k−1}$ , which has degree $(−1)^k$ . Hence deg $qϕ = deg \mathbb I+deg(−\mathbb I) = 1+(−1)^k$ , and so $d_k$ is either $0$ or multiplication by $2$ according to whether $k$ is odd or even.

Which two homeomorphisms? What happened here? It's not clear to me at all.

In addition to my queries highlighted in bold above, I also request you to please either explain step by step in the answer how to use cellular degree formula to compute $d_j$'s or direct me to some resource which explains that in great depth.

Here is my understanding of how homology groups of a Torus are computed using cellular homology: We have one zero cell $e^0$, two $1-$ cells $e_a^1, e_b^1$ and one $2-$ cell $e^2$. So $C_k= Z^2$ if $k=1$; $= Z$ if $k=0,2; =0$ else.

Computing $d_2$: It is enough to compute it on generator $e^2$. By the cellular degree formula: $d_2(e^2)= d_{a e^2} e_a^1+ d_{b e^2} e_b^1$, where $d_{a e^2}$ is the degree of the the composition $S^1\to T_1\to S^1$. The attaching map for the $2-$ cell can be denoted by $aba^{-1}b^{-1}$ (assuming the Torus to have been constructed from a square by identifying sides appropriately) and the second map collapses the complement of $a$ in $T_1$ ($1$- skeleton of the Torus) i.e., $b$ becomes a point and effectively we have $aa^{-1}$ which is a constant loop. Therefore the composition induces $H_1S^1\to H_1$ which has degree $0$ because degree of a constant map $f: S^n\to S^n$ is $0$ (why? Because by degree properties, if f is not surjective, then it has degree $0$) hence $d_{ae^2}=0$ and using exactly the same idea $d_{b e^2}=0$ so all in all $d_2(e^2)=0$.

$d_1(e_a^1)=0= d_1(e_b^1)=1=d_0(e^0)$.

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I'm going to try to answer your specifics ("why doesn't it work for $\mathbb RP^n$?") here using your/Hatcher's notation. I'll edit down the quotes a bit.

Suppose that we have chain complex for a CW comples $X$: $\cdots\to > C_{n+1}\to C_n\to C_{n-1}\to\cdots$ and let $d_j: C_j\to C_{j-1}$ be the boundary maps. Here $C_n=\mathbb Z^a, a=$ number of $n$ cells in $X$. $d_i$'s are computed by cellular boundary formula. Let me state my understanding of it: $d_j(e_\alpha^j)= \sum d_{\alpha\beta} e_\beta^{j-1},$ where $e_\alpha^j=j$ cell and $d_{\alpha\beta}=$ degree of the composition $(S^{n-1}\to X^{n-1}\to S^{n-1}),$ where the first map is the attaching map of the $n-$cell $e_\alpha^n$ and the second map is the quotient map collapsing $X^{n-1}-e_\beta^{n-1}$ to a point. I don't quite understand this second map.

To clarify (and referring to my figure-8 example in the other answer), this map is defined by thinking of the codomain, $S^{n-1}$ as the standard $n-1$-disk attached to a point, i.e., the quotient of $D^{n-1} \cup \{a\}$ by the equivalence relation in which $P \sim P$ for all $P$, and in which $P \sim a$ and $a \sim P$ for $P \in \partial D^{n-1}$. In the case of $D^1$, we take the interval $-1 \le x \le 1$, and an extra point $a$, and we identify both $+1$ and $-1$ with $a$, so that the two ends of that interval become the "same point", and we have a circle, $S^1$. Now the 'collapsing' map, which has the form $$ c: X^{n-1}\to S^{n-1} : x \mapsto \ldots $$ can be rewritten in the form $$ c: X^{n-1}\to (D^{n-1} \cup \{a\})/\sim : x \mapsto \ldots $$ with the domain, $X^{n-1}$ broken into two parts: (1) $X^{n-1}-e_\beta^{n-1}$, and (2) $e_\beta^{n-1}$, where the latter is naturally identified with $D^{n-1}$. In fact, let's let $j: D^{n-1} \to X^{n-1}$ be the mapping from the standard disk to our $(n-1)$-cell, so that $j$ is a bijection on the interior of $D^{n-1}$.
So we can now write out the map $c$: $$ c: X^{n-1} \to (D^{n-1} \cup \{a\})/ \sim : x \mapsto \begin{cases} a & x \in X^{n-1}-e_\beta^{n-1} \\ a & x \in \partial (e_\beta^{n-1}) \\ j^{-1}(x) & x \in \partial j(D^{n-1}) \end{cases} $$

This is because on page $141$, Hatcher defines $d_{\alpha\beta}$ as degree of the composition $q_\beta qq_\alpha$, where

$a) q: X^{n-1}\to X^{n-1}/X^{n-2}$ is the quotient map.

$b)q_\beta: X^{n-1}/X^{n-2}\to S_\beta^{n-1}$ is the map collapsing the complement of $e_\beta^{n-1}$ to a point. I don't understand what this map does to rest of the space.

$c) \phi_\alpha: S^{n-1}\to X^{n-1}$ is the attaching map.

I don't understand how this definition of $d_{\alpha \beta}$ matches with the earlier definition.

For now, let's assume that somehow they are same and try to compute the homology groups of $RP^n$. I know that as a cell complex it has one $k-$ cell for $k\le n$. So I get a chain complex: $\cdots \to 0\to > C_n\to C_{n-1}\to $, where $C_k=\mathbb Z$ for all $k\le n$. Now the task is to compute $d_k: C_k\to C_{k-1}$.

Here is what Hatcher does:

the attaching map for $e^k$ is the $2$ sheeted covering projection $ϕ: S^{k−1}→RP^{k−1}$. To compute the boundary map $d_k$ we compute the degree of the composition $S^{k−1}\overbrace{\to}^{\phi} RP^{k−1}\overbrace{\to}^q RP^{k−1}/RP^{k−2} = S^{k−1}$ , with $q$ the quotient map. The map $qϕ$ restricts to a homeomorphism from each component of $S^{k−1} − S^{k−2}$ onto $RP^{k−1} −RP^{k−2}$ , and these two homeomorphisms are obtained from each other by precomposing with the antipodal map of $S^{k−1}$ , which has degree $(−1)^k$ . Hence deg $qϕ = deg \mathbb I+deg(−\mathbb I) = 1+(−1)^k$ , and so $d_k$ is either $0$ or multiplication by $2$ according to whether $k$ is odd or even.

Which two homeomorphisms? What happened here? It's not clear to me at all.

Hatcher says "The map $qϕ$ restricts to a homeomorphism from each component of $S^{k−1} − S^{k−2}$ onto $RP^{k−1} −RP^{k−2}$". The space $S^{k−1} − S^{k−2}$ has two components (think of the northern and southern hemisphere of the earth, where the equator has been removed); each component is a ($k-1$)-disk. The map $qϕ$, restricted to either one of these disks, is a homeomorphism onto its image, which happens to be exactly $RP^{k−1} −RP^{k−2}$.

It might help to draw this out in the case where $k = 2$, so that you're talking about the top and bottom arcs of $S^1$ both mapping to $\mathbb RP^1$, which is the $S^1$ obtained by identifying the two endpoints of the upper arc of a circle. In that simple case, the map from the interior of upper arc of $S^1$ to $RP^1 - RP^0$ is the identity, while the map from the lower arc of $S^1$ to $RP^1 - RP^0$ involves going from $(x, y)$ to $(-x, -y)$ and THEN following the first map. This extra step is an orientation preserving map from the lower arc to the upper arc. If you try to do the same thing one dimension higher, the extra step is orientation-reversing, which is what Hatcher's getting at in the remaining bold-faced sentences.

In addition to my queries highlighted in bold above, I also request you to please either explain step by step in the answer how to use cellular degree formula to compute $d_j$'s or direct me to some resource which explains that in great depth.

Here is my understanding of how homology groups of a Torus are computed using cellular homology: We have one zero cell $e^0$, two $1-$ cells $e_a^1, e_b^1$ and one $2-$ cell $e^2$. So $C_k= Z^2$ if $k=1$; $= Z$ if $k=0,2; =0$ else.

Computing $d_2$: It is enough to compute it on generator $e^2$. By the cellular degree formula: $d_2(e^2)= d_{a e^2} e_a^1+ d_{b e^2} > e_b^1$, where $d_{a e^2}$ is the degree of the the composition $S^1\to > T_1\to S^1$. The attaching map for the $2-$ cell can be denoted by $aba^{-1}b^{-1}$ (assuming the Torus to have been constructed from a square by identifying sides appropriately) and the second map collapses the complement of $a$ in $T_1$ ($1$- skeleton of the Torus) i.e., $b$ becomes a point and effectively we have $aa^{-1}$ which is a constant loop. Therefore the composition induces $H_1S^1\to H_1$ which has degree $0$ because degree of a constant map $f: S^n\to S^n$ is $0$ (why? Because by degree properties, if f is not surjective, then it has degree $0$) hence $d_{ae^2}=0$ and using exactly the same idea $d_{b e^2}=0$ so all in all $d_2(e^2)=0$.

$d_1(e_a^1)=0= d_1(e_b^1)=1=d_0(e^0)$.

Your computation for the torus looks fine to me.

Let's look at $RP^2$. The chain groups are $$ 0 \to Z \to Z \to Z \to 0, $$ with generators (from left to right) $e^2, e^1, e^0$. If we think of a "picture" of $RP^2$ consisting of $A$, the interior upper hemisphere of $S^2$ (all points with $z > 0$, say), $B$, the "front" arc of the equator (points $(x, y, 0)$ with $y > 0$, and $C$, the rightmost point of that arc (i.e., $(1, 0, 0)$), then there's one point in our "picture" for each point of $RP^2$. The generator $e^2$ maps the interior of a 2-disk to $A$; $e^1$ maps the interior of the unit interval to $B$, and $e^2$ maps a single point to $C$, i.e., the point $(1,0,0)$.

Now $e^2$ maps the boundary of the 2-disk to the whole equator, but the points on the front arc and the back arc are identified by $(x, y, 0) \sim (-x, -y, 0)$, so you can think of the image of the 2-disk boundary as wrapping along the front arc, $B$, twice. In other words, the boundary map from $C_2$ to $C_1$, in terms of the generators $e^2$ and $e^1$, sends $e^2$ to $2e^1$.

I don't really have a good sense of why the map from $C_1$ to $C_0$ sends $e^1$ to $0e^0$ (it's hard for me to think of orientation for 0-manifolds). But this leaves us with $$ \newcommand{\ZZ}{\mathbb Z} $$ $$ 0 \to \ZZ(e^2) \overset{e^2 \mapsto 2e^1}{\longrightarrow} \ZZ(e^1) \overset{e^2 \mapsto 0e^0}{\longrightarrow} \ZZ(e^0) \to 0 $$ where $\ZZ(e^2)$ is my shorthand for 'an infinite cyclic group with a generator named $e^2$'. From this, I assume you can compute the homology groups.

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  • $\begingroup$ Thank you so much for your help and time. I'll go through the answer. $\endgroup$
    – Koro
    Sep 1, 2023 at 14:57
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Partial answer:

Let's look at a simple case: $X$ is a figure 8 with one 2-disk attached to each "circle" of the figure 8. There's a 0-cell (at the center of the figure 8), two one cells (the "loops" of the figure-8), and two 2-cells. Let's look at the case $n = 1$, and the "top" loop (I'm imagining the $8$ appearing as it does on this page).

(a) The map from $X^1$ to $X^1 / X^0$ is easy, since $X^0$ is a single point: in this case, the map is the identity.

(b) Let's pick $e^2_\beta$ to be the disk attached to the top loop, and call it $e^2_1$, and call the disk on the bottom loop $e^2_2$, OK? Then $$ q_\beta: X^{n-1}/X^{n-2}\to S_\beta^{n-1}$$ is the map $$ q_1: \text{figure-8} \to S_1^{1} $$ collapsing the complement of $e^1_1$ to a point, i.e., the map that leaves the upper loop of the figure-8 intact, and sends everything in the lower loop to the $0$-cell. If we suppose that the $0$-cell is at the origin, and everything in sight is embedded in $\mathbb R^2$, then this map is just $$ (x, y) \mapsto (x, \max(y, 0)). $$

Now that you understand what those maps look like in a specific case, does it help at all?

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  • $\begingroup$ +1. Thanks. Yes, it does. How about using this idea in $RP^{n}$? Why doesn't it seem to work? Thanks. $\endgroup$
    – Koro
    Aug 28, 2023 at 13:22
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We don't really need to understand what $q\phi$ or $\Delta_{\alpha\beta}$ or whatever "actually looks like". The local degree formula is your friend.

$\Bbb RP^n$ is a space. There is a certain cell structure on that space, suitable maps $\Phi_j:D^j\to\Bbb RP^n$ for $j=0,1,2,\cdots,n$ (just one for each $j$) which satisfy the CW axioms. We may define $\Phi_j$ as $D^j\overset{\text{north hemisphere}}{\hookrightarrow}S^j\hookrightarrow S^n\twoheadrightarrow\Bbb RP^n$.

Fix $n\ge2$. For $0\le k\le n$. The $k$ skeleton of $\Bbb RP^n$ is the union of all $\Phi_i(D^i)$, $i=0,\cdots,k$ and this is evidently the image of $S^k\hookrightarrow S^n\twoheadrightarrow\Bbb RP^n$. Fix $2\le j\le n$.

Consider that $\phi_j$ (the restriction of $\Phi_j$ to $S^{j-1}$) is just $S^{j-1}\hookrightarrow S^n\twoheadrightarrow\Bbb RP^n$. For any choice of quotient map $\Bbb RP^n_{j-1}\twoheadrightarrow S^{j-1}$ picking out the cell corresponding to $\Phi_j$ - and it's true that Hatcher isn't very careful with this - we first kill off $\Bbb RP^n_{j-2}$ and this is exactly the image $\phi_j(S^{j-2})$. Let's call the resultant map - whose degree we care about - $\Delta:S^{j-1}\to S^{j-1}$. If you pick a point $x$ of $S^{j-1}$ which is not the quotient point, it has an open neighbourhood $U$ avoiding the quotient point and $\Delta^{-1}(U)$ is just some neighbourhood $V$ in the North hemisphere of $S^{j-1}$ union its antipode $-V$ in the South hemisphere. We have that the map $H_n(-V;-V\setminus\{-p\})\to H_n(S^{j-1};S^{j-1}\setminus\{x\})$ equals the map $H_n(-V;-V\setminus\{-p\})\overset{r}{\to} H_n(V;V\setminus\{p\})\to H_n(S^{j-1};S^{j-1}\setminus\{x\})$ where $r$ is induced by $z\mapsto -z$ and we know that $r$ has degree $(-1)^j$.

Because $\phi$ is a homeomorphism on each open hemisphere and the images under $\phi$ avoid the $(j-2)$ skeleton (and all the other $(j-1)$-cells, trivially, since there is only one) we know $\Delta$ is a homeomorphism near $p$ and $-p$. So the local degree at $p$ is some number $\sigma\in\{-1,+1\}$. We don't know or care what the actual value of $\sigma$ is. Because of that factoring through $r$, we know the local degree of $\Delta$ at $-p$ is $(-1)^j\sigma$. Using the degree formula, it follows the degree of $\Delta$ is $(1+(-1)^j)\sigma$ which is zero when $j$ is odd and is $\pm2$ (we don't care which) when $j$ is even.

The first boundary map is just the simplicial homology map and is zero here since there is only one $0$-cell and the space is path connected.

So the homology is computed by the chain complex: $$0\leftarrow\Bbb Z\overset{0}{\leftarrow}\Bbb Z\overset{\pm2}{\leftarrow}\Bbb Z\overset{0}{\leftarrow}\Bbb Z\overset{\pm2}{\leftarrow}\Bbb Z\cdots\leftarrow\Bbb Z\leftarrow0\leftarrow0\cdots$$Where the last map from $\Bbb Z$ is $0$ or $\pm2$ depending on the parity of $n$. The homology of $\Bbb RP^n$ easily follows. Since $\Bbb Z\overset{(-1)}{\longrightarrow}\Bbb Z$ is an isomorphism, we can change all the signs in the chain complex to be positive or negative or alternative or whatever else you please - this does not affect homology. It is not interesting what the "actual" chain complex is.

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