Math Problems Vault

Problem 6.5.1

Suppose $(C_k, \partial_k^C)$ , $(D_k, \partial_k^D)$ are chain complexes. A sequence of homomorphisms $h_k : C_k \to D_k$ is called a chain map if $\partial_k^D h_k = h_{k-1} \partial_k^C$ . An important fact about chain maps is that they induce maps on homology by the formula $h_*([c]) = [h(c)]$ , where $[c]$ denotes the homology element represented by the chain $c$ with $\partial (c) = 0$ .

Verify this claim: show that if $[c] = [c'] \in H_n(C)$ , then $[h_n(c)] = [h_n(c')] \in H_n(D)$ .

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Problem 6.5.2

Show that if $A \xrightarrow {f} B \xrightarrow {g} C$ is a composition of chain maps (as defined in Exercise 6.5.1), then $(gf)_* = g_* f_*$ .

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Problem 6.5.3

Suppose $C, D, E$ are abelian groups with homomorphisms $f : C \to D$ and $g : D \to E$ . The sequence $C \xrightarrow {f} D \xrightarrow {g} E$ is said to be exact at $D$ if $\ker (g) = \mathrm{im}(f)$ . The sequence $0 \to C \xrightarrow {f} D \xrightarrow {g} E \to 0$ is called a short exact sequence if it is exact at $C, D, E$ ; this means that $f$ is an injection and $g$ is a surjection, as well as $\ker (g) = \mathrm{im}(f)$ .

A short exact sequence is called split if the middle term is (up to a commutative diagram of isomorphisms) a direct sum of the first and third terms, with the maps having the standard form $F(c) = (f(c), 0)$ , $G(c, e) = e$ . For example, $0 \to \mathbb {Z} \xrightarrow {f} \mathbb {Z} \oplus \mathbb {Z} \xrightarrow {g} \mathbb {Z} \to 0$ , where $f(n) = (n,0)$ , $g(n,m) = m$ , is split, while $0 \to \mathbb {Z} \xrightarrow {f} \mathbb {Z} \xrightarrow {g} \mathbb {Z}_2 \to 0$ , where $f(n) = 2n$ and $g(n) = n \bmod 2$ , is a short exact sequence that is not split (the presence of torsion in the third term obstructs splitting).

Show that if $E \simeq \mathbb {Z} \oplus \cdots \oplus \mathbb {Z}$ is a finitely generated free abelian group, then a short exact sequence $0 \to C \xrightarrow {f} D \xrightarrow {g} E \to 0$ is equivalent to a split short exact sequence (i.e. there is a commutative diagram with vertical isomorphisms connecting it to $0 \to C \xrightarrow {F} C \oplus E \xrightarrow {G} E \to 0$ ).

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Remark.

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Define a map $r : E \to D$ with $gr(e) = e$ by first defining it on generators and extending linearly. Then define a map $I : C \oplus E \to D$ by $I(c,e) = f(c) + r(e)$ . Show that $I$ gives an isomorphism and leads to the required commutative diagram. To show that $I$ is onto, start with $d \in D$ and look at $d - r(g(d))$ and show that there is $c \in C$ with $d - r(g(d)) = f(c)$ .

Problem 6.5.4

Two short exact sequences $0 \to C \xrightarrow {f} D \xrightarrow {g} E \to 0$ and $0 \to C' \xrightarrow {f’} D' \xrightarrow {g’} E' \to 0$ are called equivalent if there are vertical isomorphisms $\gamma : C \to C'$ , $\delta : D \to D'$ , $\epsilon : E \to E'$ so that the resulting diagram is commutative (i.e. $\delta f = f' \gamma$ and $\epsilon g = g' \delta$ ).

Suppose that there is a commutative diagram with exact rows,

\begin{array}{ccccccccc} 0 & \to & C & \xrightarrow {f} & D & \xrightarrow {g} & E & \to & 0 \\ & & \downarrow {\gamma } & & \downarrow {\delta } & & \downarrow {\epsilon } & & \\ 0 & \to & C’ & \xrightarrow {f’} & D’ & \xrightarrow {g’} & E’ & \to & 0 \end{array}

Show that if $\gamma$ and $\epsilon$ are isomorphisms, then $\delta$ is an isomorphism and the two short exact sequences are equivalent.

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Remark.

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The type of argument used is called diagram chasing. You have to show that $\delta$ is both surjective and injective. For surjectivity, start with $d' \in D'$ . Take $g'(d') \in E'$ and use surjectivity of $\epsilon$ to find $e \in E$ with $\epsilon (e) = g'(d')$ . Use surjectivity of $g$ to find $d \in D$ with $g(d) = e$ . Then look at $d' - \delta (d)$ . If it were $0$ , then $\delta$ would be surjective. It does not have to be $0$ , but you can use commutativity of the diagram to show that $g'(d' - \delta (d)) = 0$ . Then use exactness at $D'$ of the bottom row to show that there is $c' \in C'$ with $f'(c') = d' - \delta (d)$ . Use surjectivity of $\gamma$ to find $c$ with $\gamma (c) = c'$ . Then use commutativity of the diagram to show that $d' - \delta (d) = \delta (f(c))$ , from which you can conclude that $d' = \delta (d + f(c))$ . Then argue that $\delta$ is injective by a similar diagram chase.

Problem 6.5.5

Continuing the setup of the remark below (the connecting homomorphism $\partial : H_k(E) \to H_{k-1}(C)$ ): show that this definition is well defined. This means that if we choose another representative $e'$ of $[e] \in H_k(E)$ and go through similar steps to define $[c'] \in H_{k-1}(C)$ , then $[c'] = [c]$ .

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Remark.

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Suppose there is a short exact sequence $0 \to C \xrightarrow {f} D \xrightarrow {g} E \to 0$ of chain complexes, meaning that for each $i \geq 0$ , the sequence $0 \to C_i \xrightarrow {f} D_i \xrightarrow {g} E_i \to 0$ is a short exact sequence of abelian groups. Theorem 6.5.1 states that there is then a corresponding long exact sequence in homology,

\cdots \to H_k(C) \xrightarrow {f_*} H_k(D) \xrightarrow {g_*} H_k(E) \xrightarrow {\partial } H_{k-1}(C) \xrightarrow {f_*} \cdots ,

continuing to the right until it ends with $H_0(E) \to 0$ .

The induced maps $f_*, g_*$ are defined as in Exercise 6.5.1: $f_*([c]) = [f(c)]$ . The boundary map $\partial : H_k(E) \to H_{k-1}(C)$ is defined as follows. For an element $[e] \in H_k(E)$ , select a representative $e \in E_k$ (so $\partial e = 0$ ). Since $g$ is surjective, there is $d \in D_k$ with $g(d) = e$ . Then $g(\partial (d)) = \partial (g(d)) = \partial (e) = 0$ , so since the sequence is exact, there is $c \in C_{k-1}$ with $f(c) = \partial (d)$ . Moreover $f(\partial (c)) = \partial (f(c)) = \partial \partial (d) = 0$ , and since $f$ is injective, this implies $\partial (c) = 0$ , so $c$ represents a homology class. Define $\partial ([e]) = [c]$ .

Hint for this exercise: start with $e - e' = \partial \epsilon$ . Write $\epsilon = g(\delta )$ . Then show $g(d - d' - \partial \delta ) = 0$ and so $d - d' - \partial \delta = f(\gamma )$ . Chase around the diagram to show that $c - c' = \partial \gamma$ and so $[c] = [c']$ .

Problem 6.5.6

Continuing the setup of the remark to Exercise 6.5.5 (the long exact sequence of Theorem 6.5.1, and the connecting homomorphism $\partial$ ): show that the sequence is exact at $H_k(C)$ .

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Remark.

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This requires you to show that $\ker (f_*) = \mathrm{im}(\partial )$ . Doing so requires showing inclusions in both directions. To show that $\mathrm{im}(\partial ) \subset \ker (f_*)$ , use the fact that, if $[c] = \partial ([e])$ , then we can assume $f(c) = \partial (d)$ with $g(d) = e$ . For the other direction, if $f_*([c]) = 0$ , then $f(c) = \partial d$ . Use $d$ to find $e$ with $\partial (e) = 0$ and $\partial ([e]) = [c]$ .

Problem 6.5.7

Continuing the setup of the remark to Exercise 6.5.5: show that the sequence is exact at $H_k(D)$ .

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Remark.

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This requires you to show that $\ker (g_*) = \mathrm{im}(f_*)$ . For one direction, you just need to use $gf = 0$ . For the other direction, suppose that $g_*([d]) = 0$ , which means that $g(d) = \partial (\epsilon )$ . Choose $\delta$ so that $g(\delta ) = \epsilon$ . Then show that $g(d - \partial (\delta )) = 0$ and use this to show that there is $c$ with $\partial (c) = 0$ and $f_*([c]) = [d]$ .

Problem 6.5.8

Continuing the setup of the remark to Exercise 6.5.5: show that the sequence is exact at $H_k(E)$ .

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Remark.

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This requires you to show that $\ker (\partial ) = \mathrm{im}(g_*)$ . To show that $\ker (\partial ) \subset \mathrm{im}(g_*)$ , suppose $[e] \in H_k(E)$ satisfies $\partial ([e]) = 0$ . Then there are elements $d, c, \gamma$ with $g(d) = e$ , $f(c) = \partial (d)$ , $c = \partial (\gamma )$ . Then show that $\partial (d - f(\gamma )) = 0$ and $g_*([d - f(\gamma )]) = [e]$ .

Problem 6.5.9

Prove Theorem 6.5.2 (naturality of the long exact sequence): suppose that there are two short exact sequences of chain complexes which are connected via chain maps $\alpha , \beta , \gamma$ , represented in the commutative diagram

\begin{array}{ccccccccc} 0 & \to & C & \xrightarrow {f} & D & \xrightarrow {g} & E & \to & 0 \\ & & \downarrow {\alpha } & & \downarrow {\beta } & & \downarrow {\gamma } & & \\ 0 & \to & C’ & \xrightarrow {f’} & D’ & \xrightarrow {g’} & E’ & \to & 0 \end{array}

Show that the corresponding long exact sequences (Theorem 6.5.1) are connected via the commutative diagram

\begin{array}{ccccccccccc} \cdots \to & H_k(C) & \xrightarrow {f_*} & H_k(D) & \xrightarrow {g_*} & H_k(E) & \xrightarrow {\partial } & H_{k-1}(C) & \to & \cdots \\ & \downarrow {\alpha _*} & & \downarrow {\beta _*} & & \downarrow {\gamma _*} & & \downarrow {\alpha _*} & & \\ \cdots \to & H_k(C’) & \xrightarrow {f_*’} & H_k(D’) & \xrightarrow {g_*’} & H_k(E’) & \xrightarrow {\partial } & H_{k-1}(C’) & \to & \cdots \end{array}

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Problem 6.5.10

Show that if there is a long exact sequence of abelian groups,

A \xrightarrow {f} B \xrightarrow {g} C \xrightarrow {h} D \xrightarrow {k} E,

then there is a corresponding short exact sequence

0 \to \mathrm{coker}(f) \xrightarrow {\bar{g}} C \xrightarrow {\bar{h}} \ker (k) \to 0,

where $\mathrm{coker}(f) = B/\mathrm{im}(f)$ .

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Problem 6.5.11

(a)(b)(c)

Suppose there is a commutative diagram of abelian groups with exact rows,

\begin{array}{ccccccccccc} A & \xrightarrow {f} & B & \xrightarrow {g} & C & \xrightarrow {h} & D & \xrightarrow {k} & E \\ \downarrow {\alpha } & & \downarrow {\beta } & & \downarrow {\gamma } & & \downarrow {\delta } & & \downarrow {\epsilon } \\ A’ & \xrightarrow {f’} & B’ & \xrightarrow {g’} & C’ & \xrightarrow {h’} & D’ & \xrightarrow {k’} & E’ \end{array}

where $\beta , \delta$ are isomorphisms, $\alpha$ is surjective, and $\epsilon$ is injective.

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(a)

Show that this diagram induces a commutative diagram

\begin{array}{ccccccccc} 0 & \to & \mathrm{coker}(f) & \xrightarrow {\bar{g}} & C & \xrightarrow {\bar{h}} & \ker (k) & \to & 0 \\ & & \downarrow {\bar{\beta }} & & \downarrow {\gamma } & & \downarrow {\bar{\delta }} & & \\ 0 & \to & \mathrm{coker}(f’) & \xrightarrow {\bar{g}’} & C’ & \xrightarrow {\bar{h}’} & \ker (k’) & \to & 0 \end{array}

(using the short exact sequence associated to a long exact sequence, Exercise 6.5.10).

(b)

Show that $\bar{\beta }$ and $\bar{\delta }$ are isomorphisms.

(c)

Use Exercise 6.5.4 to show that $\gamma$ is an isomorphism.

Problem 6.5.12

Chain maps $F, G : C \to D$ are called chain homotopic if there is a map $H_k : C_k \to D_{k+1}$ for each $k \geq 0$ so that $G_k - F_k = \partial_{k+1}^D H_k + H_{k-1} \partial_k^C$ .

Show that if the chain maps $F, G$ are chain homotopic, then $F$ and $G$ induce the same map on homology. This involves showing that if $c$ is a cycle in $C_k$ , then $G_k(c) - F_k(c)$ is a boundary.

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