Math Problems Vault

Problem 6.10.1

(a)(b)

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(a)

Show the inclusion $(D^k, S^{k-1}) \to (D^k, L^{k-1})$ induces an isomorphism in homology.

(b)

Show that the inclusion $(X^k, X^{k-1}) \to (X^k, K^{k-1})$ induces an isomorphism in homology.

Remark.

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For part (a): show that there is a homotopy equivalence of pairs. For part (b): use a similar argument.

Setup: $X^k$ denotes the $k$ -skeleton of $X$ , the union of the cells of dimension $\leq k$ . The $k$ -cell $e_j^k$ is a unit disk (up to homeomorphism) attached to $X^{k-1}$ via an attaching map $f_j : S^{k-1} \to X^{k-1}$ . Extend $X^{k-1}$ to a larger set $K^{k-1}$ in $X^k$ by attaching $L^{k-1} = D^k \setminus (\mathrm{int}\, \frac{1}{2} D^k)$ via $f_j$ for each $k$ -cell. There is a corresponding pair $(D_j^k, L_j^{k-1})$ for each $k$ -cell; write $(D^k, L^{k-1}) = (\bigsqcup_{j=1}^{c_k} D_j^k, \bigsqcup_{j=1}^{c_k} L_j^{k-1})$ and $(D^k, S^{k-1}) = (\bigsqcup_{j=1}^{c_k} D_j^k, \bigsqcup_{j=1}^{c_k} S_j^{k-1})$ for the corresponding disjoint unions.

Problem 6.10.2

(a)(b)(c)

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(a)

Show that if $(Y, B) = (\bigsqcup_{j=1}^N Y_j, \bigsqcup_{j=1}^N B_j)$ is a disjoint union of path components $(Y_j, B_j)$ , then $H_p(Y, B) = \bigoplus_{j=1}^N H_p(Y_j, B_j)$ .

(b)

Show that $H_k(D^k, S^{k-1}) \simeq \bigoplus_{j=1}^{c_k} \mathbb {Z}$ (notation as in Exercise 6.10.1). Call the generator $g_j^k$ that corresponds to a generator of $H_k(D_j^k, S_j^{k-1})$ .

(c)

Show that the inclusion $H_p(D^k, S^{k-1}) \to H_p(D^k, L^{k-1})$ induces an isomorphism in homology. Conclude that

H_p(D^k, L^{k-1}) = \bigoplus _{j=1}^{c_k} H_p(D_j^k, L_j^{k-1}) \simeq \bigoplus _{j=1}^{c_k} \mathbb {Z}.

Problem 6.10.3

Show that the restriction of the characteristic maps to the $k$ -cells (notation as in Exercise 6.10.1) gives a homeomorphism of pairs $(D^k \setminus \mathrm{int}\, L^{k-1}, L^{k-1} \setminus \mathrm{int}\, L^{k-1}) \to (X^k \setminus \mathrm{int}\, K^{k-1}, K^{k-1} \setminus \mathrm{int}\, K^{k-1})$ . Conclude that there is an isomorphism in homology.

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Problem 6.10.4

Show that there is a commutative diagram where all maps are isomorphisms,

\begin{array}{ccc} H_p(D^k, S^{k-1}) & \xrightarrow {\phi _*} & H_p(X^k, X^{k-1}) \\ \downarrow {(i_L)_*} & & \downarrow {(i_K)_*} \\ H_p(D^k, L^{k-1}) & \xrightarrow {(\phi _L)_*} & H_p(X^k, K^{k-1}) \\ \uparrow {(j_L)_*} & & \uparrow {(j_K)_*} \\ H_p(D^k \setminus \mathrm{int}\, L^{k-1}, L^{k-1} \setminus \mathrm{int}\, L^{k-1}) & \xrightarrow {(\phi _{-L})_*} & H_p(X^k \setminus \mathrm{int}\, K^{k-1}, K^{k-1} \setminus \mathrm{int}\, K^{k-1}) \end{array}

where $\phi = \sqcup \phi_j$ is the disjoint union of the characteristic maps of the $k$ -cells (restricting to the attaching maps on the boundary spheres), the other horizontal maps are induced by restrictions of this map, and the vertical maps are inclusions. Use Exercises 6.10.1--6.10.3 to justify why each map is an isomorphism.

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Problem 6.10.5

Show that when $k > 0$ , $H_k(\bigvee_{j=1}^{c_k} S_j^k, *) \simeq \bigoplus_{j=1}^{c_k} H_k(S_j^k, *)$ .

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Remark.

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Combining the results of Exercises 6.10.3 and 6.10.4 shows that the characteristic maps induce an isomorphism $\bigoplus_{j=1}^{c_k} H_p(D_j^k, S_j^{k-1}) \to H_p(X^k, X^{k-1})$ ; in particular all terms are $0$ unless $p = k$ , and when $p = k$ the generator $g_j^k$ of $H_p(D_j^k, S_j^{k-1})$ is mapped to a generator $x_j^k$ of $H_k(X^k, X^{k-1})$ , which is thus a free abelian group with generators $x_1^k, \ldots , x_{c_k}^k$ .

For $k > 0$ , the quotient space $D^k/S^{k-1}$ is homeomorphic to $S^k$ . Starting with the disjoint union $(D^k, S^{k-1})$ and forming the quotient $D^k/S^{k-1}$ gives the space formed from $c_k$ copies of $S^k$ with all the base points (coming from the $S_j^{k-1}$ terms collapsed to a point) identified to a single base point $*$ ; thus the quotient space is $\bigvee_{j=1}^{c_k} S_j^k$ . The map $\phi$ (of Exercise 6.10.4) induces a homeomorphism $\bar{\phi }$ between $X^k/X^{k-1}$ and this wedge, giving an isomorphism $H_p(\bigvee_{j=1}^{c_k} S_j^k, *) \simeq H_p(X^k/X^{k-1}, *)$ .

Hint for this exercise: use an inductive argument with the Mayer--Vietoris sequence.

Problem 6.10.6

Any element $y$ of the direct sum $\bigoplus_{j=1}^{c_k} H_k(S_j^k, *)$ of Exercise 6.10.5 is expressible as $\sum_{j=1}^{c_k} n_j y_j$ , where $y_j$ is a chosen generator of $H_k(S_j^k, *)$ . Let $A_j$ represent the spheres other than the $j$ th one in $\bigvee_{j=1}^{c_k} S_j^k$ ; the quotient $\bigvee_{j=1}^{c_k} S_j^k / A_j$ is homeomorphic to $S_j^k$ , and under the quotient map $P_j$ collapsing $A_j$ , $y$ maps to $n_j y_j$ . This uses the fact that the composition $S_j^k \to \bigvee_{j=1}^{c_k} S_j^k \to \bigvee_{j=1}^{c_k} S_j^k/A_j = S_j^k$ is the identity, and that the composition $S_p^k \to \bigvee_{j=1}^{c_k} S_j^k \to \bigvee_{j=1}^{c_k} S_j^k/A_j = S_j^k$ maps $S_p^k$ to the base point $*$ when $p \neq j$ .

Verify that the map $S_p^k \to S_j^k$ , $p \neq j$ , which sends the whole sphere to the base point, induces the zero map in homology in dimension $k > 0$ (this is the fact used above to conclude that the identity induces the identity map while the constant map induces the zero map, giving the algebraic description of $P_j$ ).

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Problem 6.10.7

Let $A_j \subset D^k$ , $B_j \subset X^k$ denote the subcomplexes consisting of everything except the $j$ th $k$ -cell (notation as in Exercise 6.10.1). We can form quotient spaces $D^k/A_j$ , $X^k/B_j$ by collapsing $A_j$ and $B_j$ to a point. Show that the characteristic map $D_j^k \to D^k \to X^k$ induces homeomorphisms between the quotient spaces $D_j^k/S_j^{k-1} \to D^k/A_j \to X^k/B_j$ .

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Problem 6.10.8

Each sphere $S_j^k$ (as in Exercise 6.10.5) has a closed neighborhood $\bar{L}_j^{k-1}$ about the base point $*$ , coming from the image of $L_j^{k-1}$ under the collapsing map. The union of all of these in $\bigvee_{j=1}^{c_k} S_j^k$ gives a closed neighborhood $\bar{L}^{k-1}$ of $*$ . Similarly, $K^{k-1}$ (as in Exercise 6.10.1) provides a closed neighborhood $\bar{K}^{k-1}$ of the base point $* \in X^k/X^{k-1}$ ; the homeomorphism $\bar{\phi }$ (Exercise 6.10.5) sends $\bar{L}^{k-1}$ to $\bar{K}^{k-1}$ .

Show that $(D^k/S^{k-1}, \bar{L}^{k-1})$ is homotopy equivalent to $(D^k/S^{k-1}, *)$ , and $(X^k/X^{k-1}, \bar{K}^{k-1})$ to $(X^k/X^{k-1}, *)$ .

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Problem 6.10.9

Show that the restriction of the quotient map gives a homeomorphism

(X^k \setminus \mathrm{int}\, K^{k-1}, K^{k-1} \setminus \mathrm{int}\, K^{k-1}) \to (X^k/X^{k-1} \setminus \mathrm{int}\, \bar{K}^{k-1}, \bar{K}^{k-1} \setminus \mathrm{int}\, \bar{K}^{k-1}).

(notation as in Exercise 6.10.8).

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Problem 6.10.10

(a)(b)(c)

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(a)

Suppose $k > 0$ . Show that there is a commutative diagram where all maps are isomorphisms, horizontal maps induced by inclusions and vertical maps induced by quotient maps,

\begin{array}{ccccc} H_k(X^k, X^{k-1}) & \to & H_k(X^k, K^{k-1}) & \leftarrow & H_k(X^k \setminus \mathrm{int}\, K^{k-1}, K^{k-1} \setminus \mathrm{int}\, K^{k-1}) \\ \downarrow & & \downarrow & & \downarrow \\ H_k(X^k/X^{k-1}, *) & \to & H_k(X^k/X^{k-1}, \bar{K}^{k-1}) & \leftarrow & H_k(X^k/X^{k-1} \setminus \mathrm{int}\, \bar{K}^{k-1}, \bar{K}^{k-1} \setminus \mathrm{int}\, \bar{K}^{k-1}) \end{array}

(b)

Deduce a similar result for $(D^k, S^{k-1})$ in place of $(X^k, X^{k-1})$ .

(c)

Show that there is a commutative diagram, where all maps are isomorphisms,

\begin{array}{ccc} H_k(D^k, S^{k-1}) & \to & H_k(X^k, X^{k-1}) \\ \downarrow & & \downarrow \\ H_k(D^k/S^{k-1}, *) & \to & H_k(X^k/X^{k-1}, *) \\ & \searrow & \downarrow \\ & & H_k(\bigvee _{j=1}^{c_k} S_j^k, *) \end{array}

Problem 6.10.11

Define a generator $y_j^k$ of $H_k(\bigvee_{j=1}^{c_k} S_j^k, *)$ coming from the image of the generator $g_j^k \in H_k(D^k, S^{k-1})$ (using the commutative diagram of Exercise 6.10.10(c), this is the image of $x_j^k$ via the vertical maps on the right). If $P_j$ collapses all but the $j$ th sphere of $\bigvee_{j=1}^{c_k} S_j^k$ , then $(P_j)_*(y_j^k) = y_j^k$ and $(P_m)_*(y_j^k) = 0$ , $m \neq j$ , by Exercise 6.10.6.

Define the cellular chain groups $C_k^c(X) = H_k(X^k, X^{k-1})$ , $k \geq 0$ , where $X^{-1} = \emptyset$ , so $C_0^c(X) = H_0(X^0)$ ; this is a free abelian group with generators $x_j^k$ , one for each $k$ -cell. Define the boundary homomorphism $\partial_k^c$ to be the composition

H_k(X^k, X^{k-1}) \xrightarrow {\partial } H_{k-1}(X^{k-1}) \xrightarrow {j_*} H_{k-1}(X^{k-1}, X^{k-2}).

(When $k = 1$ , this is just $\partial_1^c = \partial : H_1(X^1, X^0) \to H_0(X^0)$ ; as usual $\partial_0^c = 0$ .)

Show that $\partial_k^c \partial_{k+1}^c = 0$ .

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Remark.

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Use the exactness of the long exact sequence of the pair $(X^k, X^{k-1})$ .

Problem 6.10.12

For $S^2 \times S^2$ , there are cells in dimensions $0, 2, 4$ , and the cellular chain complex is $0 \to \mathbb {Z} \to 0 \to \mathbb {Z} \oplus \mathbb {Z} \to 0 \to \mathbb {Z}$ in low dimensions $0$ -- $5$ . Since all boundary maps are $0$ , the homology groups are the chain groups, giving

H_k^c(S^2 \times S^2) \simeq \begin{cases} \mathbb {Z} & k = 0, 4, \\ \mathbb {Z} \oplus \mathbb {Z} & k = 2, \\ 0 & \text{otherwise}. \end{cases}

Show, by a similar computation with the cellular chain complex of $\mathbb {CP}^n$ , that

H_k^c(\mathbb {CP}^n) = \begin{cases} \mathbb {Z} & 0 \leq k \leq 2n, \, k \text{ even}, \\ 0 & \text{otherwise}. \end{cases}

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Problem 6.10.13

Compute the cellular homology of $S^1 \times S^2$ , using the cellular chain complex $(C^c_k(X), \partial_k^c)$ of Exercise 6.10.11.

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Remark.

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Worked model (the cellular homology of $S^1 \times \mathbb {RP}^2$ ): the chain complex is $C_3^c = \mathbb {Z}(x^3) \to C_2^c = \mathbb {Z}(x_1^2) \oplus \mathbb {Z}(x_2^2) \to C_1^c = \mathbb {Z}(x_1^1) \oplus \mathbb {Z}(x_2^1) \to C_0^c = \mathbb {Z} \to 0$ , where $x_1^1$ is the generator from the $1$ -cell forming the circle $S^1 \times y$ , $x_2^1$ is the generator from the $1$ -cell forming the circle $x \times S^1 \subset \mathbb {RP}^2$ , $x_1^2$ is the generator from the $2$ -cell forming the torus, $x_2^2$ is the generator from the $2$ -cell used in forming $x \times \mathbb {RP}^2$ , and $x^3$ is the generator for the $3$ -cell. Here $\partial_1^c = 0$ , and from the methods for two-dimensional complexes, $\partial_2^c(x_1^2) = 0$ , $\partial_2^c(x_2^2) = 2x_2^1$ . To compute $\partial_3^c$ , look at the attaching map of the $3$ -cell, $f : S^2 \to X^2$ ; writing $S^2 = \partial (D^1 \times D^2) = S^0 \times D^2 \cup D^1 \times S^1$ , this map sends each copy of $\left\{ \pm 1\right\} \times D^2$ via the characteristic map of the $2$ -cell in $\mathbb {RP}^2$ , and sends $D^1 \times S^1$ via the characteristic map of the first circle crossed with the attaching map of the $2$ -cell in $\mathbb {RP}^2$ (a degree- $2$ map). The two copies of $D^2$ are mapped the same way but inherit different orientations in $S^2$ , so they contribute $0$ to $\partial_3(x^3)$ (equivalently, they have opposite local degrees at the two points in the inverse image of a point); $D^1 \times S^1$ maps with a degree- $2$ map to the torus, giving $\partial_3^c(x^3) = 2x_1^2$ . Thus $\ker (\partial_3) = 0$ , so $H_3^c(S^1 \times \mathbb {RP}^2) = 0$ ; $\ker (\partial_2^c)$ is generated by $x_1^2$ and $\mathrm{im}(\partial_3)$ by $2x_1^2$ , so $H_2(S^1 \times \mathbb {RP}^2) \simeq \mathbb {Z}_2$ ; $\mathrm{im}(\partial_2^c)$ is generated by $2x_2^1$ , so $H_1(S^1 \times \mathbb {RP}^2) = \mathbb {Z} \oplus \mathbb {Z}_2$ ; and $H_0(S^1 \times \mathbb {RP}^2) = \mathbb {Z}$ , all other homology vanishing.

Problem 6.10.14

Compute the cellular homology of $\mathbb {RP}^2 \times \mathbb {RP}^2$ , using the cellular chain complex $(C^c_k(X), \partial_k^c)$ of Exercise 6.10.11.

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Problem 6.10.15

Compute the cellular homology of $\mathbb {RP}^n$ , using the cellular chain complex $(C^c_k(X), \partial_k^c)$ of Exercise 6.10.11.

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Remark.

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Worked model (the cellular homology of $\mathbb {RP}^3$ and $\mathbb {RP}^4$ ): the cellular chain complex of $\mathbb {RP}^3$ is $0 \to \mathbb {Z} \to \mathbb {Z} \to \mathbb {Z} \to \mathbb {Z} \to 0$ . The $2$ -skeleton is $\mathbb {RP}^2$ , with $\partial_2^c(x^2) = 2x^1$ , $\partial_1^c(x^1) = 0$ ; since $\mathrm{im}(\partial_3^c) \subset \ker (\partial_2^c) = 0$ , we must have $\partial_3^c = 0$ , giving

H_k^c(\mathbb {RP}^3) = \begin{cases} \mathbb {Z} & k = 0, 3, \\ \mathbb {Z}_2 & k = 1, \\ 0 & \text{otherwise}. \end{cases}

This can also be seen using the fact (Exercise 6.9.15) that the degree of the map $S^2 \to \mathbb {RP}^2 \to \mathbb {RP}^2/S^1 = S^2$ is $1 + (-1)^3 = 0$ . For higher real projective spaces algebra alone no longer determines the top boundary map, but Exercise 6.9.15 still applies: the attaching map of the $4$ -cell of $\mathbb {RP}^4$ is the standard double covering $S^3 \to \mathbb {RP}^3$ , composed with collapsing $\mathbb {RP}^2 \subset \mathbb {RP}^3$ to a point to get $S^3$ ; each hemisphere covers the $3$ -sphere exactly once, and the two hemisphere maps are related by $f(-x) = f(x)$ , i.e. $f = fA$ with $A(x) = -x$ . Since $A : S^3 \to S^3$ has degree $1$ , the local degree is the same at both points in the inverse image of a point, so the generator of $S^3$ is sent to twice the generator, giving $\partial^c(x^4) = \pm 2x^3$ . Hence $H_4(\mathbb {RP}^4) = 0$ , $H_3(\mathbb {RP}^4) = \mathbb {Z}_2$ , and since the $3$ -skeleton of $\mathbb {RP}^4$ agrees with $\mathbb {RP}^3$ , the lower-dimensional homology agrees too, giving

H_k(\mathbb {RP}^4) = \begin{cases} \mathbb {Z} & k = 0, \\ \mathbb {Z}_2 & k = 1, 3, \\ 0 & \text{otherwise}. \end{cases}