The Lumenary · First-year graduate

Homological algebra

Name: ______________________

Date: ______________________

You will learn to

Compute the homology of chain complexes, and derive the long exact homology sequence of a short exact sequence of complexes from the snake lemma
Prove that chain-homotopic maps induce the same map on homology, and prove the comparison theorem for projective resolutions
Define left and right derived functors, explain why they measure the failure of exactness, and compute Ext and Tor over $\Z$ and $k[x]$
Show that $\mathrm{Ext}^1(A, B)$ classifies extensions $0 \to B \to E \to A \to 0$ up to equivalence
Distinguish isomorphism, quasi-isomorphism and homotopy equivalence of complexes, with counterexamples

1.
Consider the chain complex $0 \to \mathbb{Z} \xrightarrow{d_2} \mathbb{Z}^2 \xrightarrow{d_1} \mathbb{Z} \to 0$ with $d_2(n) = (2n, 4n)$ and $d_1(a, b) = 2a - b$ . How many elements does $H_1 = \ker d_1 / \operatorname{im} d_2$ have?
2.
Which statements about chain maps are true in general? (Choose all that apply.)
- A.Every homotopy equivalence is a quasi-isomorphism
- B.Every quasi-isomorphism is a homotopy equivalence
- C.Chain-homotopic maps induce the same maps on homology
- D.A chain map that induces $0$ on all homology groups is null-homotopic
3.
How many elements does $\mathrm{Tor}_1^{\mathbb{Z}}(\mathbb{Z}/12, \mathbb{Z}/18)$ have?
4.
How many equivalence classes of extensions $0 \to \mathbb{Z}/5 \to E \to \mathbb{Z}/5 \to 0$ of abelian groups are there?
5.
In the existence part of the comparison theorem, why does $f_n d : P_{n+1} \to Q_n$ land in $\ker(d : Q_n \to Q_{n-1})$ (for $n \geq 1$ )?
- A. $d f_n d = f_{n-1} d d = 0$ , using that $f_n$ commutes with $d$ and $d \circ d = 0$
- B.Because $P_{n+1}$ is projective
- C.Because $Q_\bullet$ is exact
- D.Because the lift $f_n$ is unique
6.
Apply $- \otimes \mathbb{Z}/2$ and its derived functors to $0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z}/2 \to 0$ . The long exact sequence reads $\mathrm{Tor}_1(\mathbb{Z}, \mathbb{Z}/2) \to \mathrm{Tor}_1(\mathbb{Z}/2, \mathbb{Z}/2) \xrightarrow{\partial} \mathbb{Z}/2 \xrightarrow{2} \mathbb{Z}/2 \to \mathbb{Z}/2 \to 0$ . What is $\mathrm{Tor}_1(\mathbb{Z}/2, \mathbb{Z}/2)$ ?
- A. $\mathbb{Z}/2$
- B. $0$
- C. $\mathbb{Z}$
- D. $\mathbb{Z}/4$
7.
Which Lean statement says that the extension $0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z}/2 \to 0$ does not split (as proved in not_split)?
- A.¬ ∃ s : ZMod 2 →+ ℤ, ∀ x : ZMod 2, ((s x : ℤ) : ZMod 2) = x
- B.∃ s : ZMod 2 →+ ℤ, ∀ x : ZMod 2, ((s x : ℤ) : ZMod 2) = x
- C.¬ ∃ r : ℤ →+ ZMod 2, ∀ n : ℤ, r n = (n : ZMod 2)
- D.∀ f : ZMod 2 →+ ℤ, f ≠ 0
8.
A chain complex $0 \to C_2 \xrightarrow{d_2} C_1 \xrightarrow{d_1} C_0 \to 0$ of free abelian groups is given by integer matrices d2 (shape $\dim C_1 \times \dim C_2$ ) and d1 (shape $\dim C_0 \times \dim C_1$ ), as lists of rows. Write betti_mod_p(d2, d1, p) returning [b0, b1, b2], the dimensions of $H_n(C \otimes \mathbb{F}_p)$ for a prime $p$ , using $b_n = \dim C_n - \operatorname{rank} d_n - \operatorname{rank} d_{n+1}$ over $\mathbb{F}_p$ (with $d_0 = d_3 = 0$ ).
Computational — Practice

Answer key — Homological algebra

1.
$\ker d_1 = \{(a, b) : b = 2a\} = \mathbb{Z}(1, 2)$ and $\operatorname{im} d_2 = \mathbb{Z}(2, 4) = 2\,\mathbb{Z}(1, 2)$ , so $H_1 \cong \mathbb{Z}/2$ , with $2$ elements. Also $H_2 = \ker d_2 = 0$ and $H_0 = \mathbb{Z}/\operatorname{im} d_1 = 0$ since $d_1(0, -1) = 1$ .
2.
True: homotopy equivalences are quasi-isomorphisms, and homotopic maps agree on homology. False: quasi-isomorphisms need not have homotopy inverses ( $[\mathbb{Z} \xrightarrow{2} \mathbb{Z}] \to \mathbb{Z}/2$ ), and a map inducing zero on homology need not be null-homotopic: if $\mathrm{id} = ds + sd$ on $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0$ , then $s : \mathbb{Z}/2 \to \mathbb{Z}$ would be a splitting, and there is none.
3.
Tensoring the resolution with $\mathbb{Z}/18$ gives $\mathbb{Z}/18 \xrightarrow{12} \mathbb{Z}/18$ , whose kernel $\{x : 12x \equiv 0 \pmod{18}\} = \{0, 3, 6, 9, 12, 15\}$ has $\gcd(12, 18) = 6$ elements. So $\mathrm{Tor}_1(\mathbb{Z}/12, \mathbb{Z}/18) \cong \mathbb{Z}/6$ .
4.
$\mathrm{Ext}^1(\mathbb{Z}/5, \mathbb{Z}/5) \cong (\mathbb{Z}/5)/5(\mathbb{Z}/5) = \mathbb{Z}/5$ , so there are $5$ classes: the split one, with $E = (\mathbb{Z}/5)^2$ , and four non-split ones, all with $E \cong \mathbb{Z}/25$ but inequivalent, because they identify the quotient $\mathbb{Z}/25 / 5\mathbb{Z}/25$ with $\mathbb{Z}/5$ in different ways.
5.
By induction $d f_n = f_{n-1} d$ , so $d (f_n d) = f_{n-1} d d = 0$ . Hence $f_n d$ maps $P_{n+1}$ into $\ker(Q_n \to Q_{n-1})$ , which by exactness is the image of $Q_{n+1} \to Q_n$ ; since $P_{n+1}$ is projective, $f_n d$ lifts to $f_{n+1} : P_{n+1} \to Q_{n+1}$ with $d f_{n+1} = f_n d$ .
6.
$\mathrm{Tor}_1(\mathbb{Z}, \mathbb{Z}/2) = 0$ , so $\partial$ is injective, and its image is the kernel of $2 \otimes \mathrm{id} : \mathbb{Z} \otimes \mathbb{Z}/2 \to \mathbb{Z} \otimes \mathbb{Z}/2$ , which is the zero map $\mathbb{Z}/2 \to \mathbb{Z}/2$ . So $\mathrm{Tor}_1(\mathbb{Z}/2, \mathbb{Z}/2) \cong \mathbb{Z}/2$ , in agreement with $\mathrm{Tor}_1(\mathbb{Z}/m, \mathbb{Z}/n) \cong \mathbb{Z}/\gcd(m, n)$ . The connecting map is exactly the snake's diagonal: the kernel lost by tensoring reappears as $\mathrm{Tor}_1$ .
7.
Splitting means a homomorphism $s : \mathbb{Z}/2 \to \mathbb{Z}$ with $q \circ s = \mathrm{id}$ , i.e. ∀ x, ((s x : ℤ) : ZMod 2) = x; non-splitting is the negation of its existence. The Lean proof uses hom_zmod_two_int: every such $s$ is $0$ , and then $q(s(1)) = 0 \neq 1$ .

def rank_mod_p(matrix, p):
    rows = [[v % p for v in row] for row in matrix]
    rank = 0
    ncols = len(rows[0]) if rows else 0
    for c in range(ncols):
        piv = next((i for i in range(rank, len(rows)) if rows[i][c]), None)
        if piv is None:
            continue
        rows[rank], rows[piv] = rows[piv], rows[rank]
        inv = pow(rows[rank][c], p - 2, p)
        rows[rank] = [v * inv % p for v in rows[rank]]
        for i in range(len(rows)):
            if i != rank and rows[i][c]:
                k = rows[i][c]
                rows[i] = [(a - k * b) % p for a, b in zip(rows[i], rows[rank])]
        rank += 1
    return rank


def betti_mod_p(d2, d1, p):
    c0, c1, c2 = len(d1), len(d2), len(d2[0])
    r1, r2 = rank_mod_p(d1, p), rank_mod_p(d2, p)
    return [c0 - r1, c1 - r1 - r2, c2 - r2]

For $\mathbb{R}P^2$ the boundary $d_2 = (2, 2)^{\mathsf T}$ vanishes mod $2$ but not mod $3$ , so the $\mathbb{F}_2$ -Betti numbers are $1, 1, 1$ while the $\mathbb{F}_3$ -Betti numbers are $1, 0, 0$ — the universal coefficient theorem at work: $H_1(\mathbb{R}P^2; \mathbb{Z}) = \mathbb{Z}/2$ is seen only by coefficients of characteristic $2$ . The test with $d_2 = (0, 2)^{\mathsf T}$ is the Klein bottle.