Prove that \(Tot(C)\) is acyclic if the double complex is bounded and the rows and columns are exact.

Let \(C\) be a bounded double complex with exact rows and columns and consider \(x=(x_{p,q}) \in Tot(C)_n\), where \(p+q = n\) for any integers \(p,q\).

Recall that \(d(x_{p,q}) = d^h(x_{p,q}) + d^v(x_{p,q})\). Since \(C\) is bounded, we can look at the north eastern most nontrivial module \(C_{i,j}\) and hence nontrivial \(Tot(C)_k\).

The procedure is as follow: Start with \(d(x_{p,q}) = d^h(x_{p,q}) + d^v(x_{p,q})\), note that if \(x_{p,q}\) lies in the kernel of \(d\), then there exists an element \(x_{p-1,q+1}\) such that \(d^h(x_{p-1,q+1}) = 0\). Exactness of our columns then tells us that \(x_{p-1,q+1}=d^v(x_{p-1,q+2})\). An identical argument can be made for the rows. Combining these facts together we see that every element in \(Ker(d_{n})\) comes from an element in \(Im(d_{n-1}\). Hence \(Im(d_{n-1}) = Ker(d_n)\)

Thus \(Tot(C)\) is acyclic.

Exercise 1.3.1

Let \(0 \to A \to B \to C \to 0\) be a short exact sequence of complexes. Show that if two of the three complexes A,B,C are exact, then so is the third.

Solution: First, let \(B,C\) be exact complexes. Then, their homologies are 0 everywhere. So we can look at instead the short exact sequence \(0 \to H_n(A) \to 0 \to 0 \to 0\). Wherein exactness tells us immediately that \(H_n(A)\) is trivial, hence \(A\) is exact.

This argument can be repeated for \(A,C\) and \(A,B\). So we're done.

This one will be turned in by hand, it's better with the diagram and I can't use tikz in this environment.

Let \(f\) be a morphism of chain complexes. Show that if \(ker(f)\) and \(coker(f)\) are acyclic, then \(f\) is a quasi-isomorphism. Is the converse true?

Let \(A,B\) be chain complexes and \(f: A \to B\) a morphism. We will show that if \(ker(f)\) and \(coker(f)\) are acyclic, then \(f\) is a quasi-isomorphism.

Notice that we get a short exact sequence \(0 \to ker(f) \to A \to Im(f) \to 0\) and \(0 \to Im(f) \to B \to coker(f) \to 0\). If we look at the long exact sequence of homology that is induced from the first part. It looks like \(0 \to 0 \to H_n(A) \to H_n(Im(f)) \to 0\) since both \(ker(f)\) and \(Im(f)\) are acyclic. Exactness of this sequence tells us that \(H_n(A) \cong H_n(Im(f))\). We can make the exact same argument for the other short exact sequence and the induced long exact sequence on homology for that. Which means we get \(H_n(A) \cong H_n(Im(f)) \cong H_(B)\). Thus \(f\) is a quasi-isomorphism from A to B.

The converse is not true. This is show by the following example: Let \(A\) be the chain complex \(\ldots \to 0 \to 0 \to \mathbb{Z} \to \mathbb{Z} \to 0 \to \ldots\) and \(B\) the chain complex \(\ldots \to 0 \to \mathbb{Z} \to \mathbb{Z} \to 0 \to 0\to \ldots \) where only the middle \(\mathbb{Z}\) terms coincide, everywhere else 0.

If we let \(f\) be the chain complex morphism \(f: A \to B\) that sends the middle \(\mathbb{Z}\) to the middle \(\mathbb{Z}\) and everywhere else 0, then \(ker(f)\) and \(coker(f)\) are not acyclic. But \(f\) is a quasi-isomorphism. Since the Homologies of \(ker(f)\) and \(coker(f)\) are nontrivial at exactly one place.

1. Show that acyclic bounded below chain complexes of free R-modules are always split exact.

1. Show that an acyclic chain complex of finitely generated free abelian groups is always split exact, even when it is not bounded below.

Let \(A.\) be an acyclic bounded below chain complex of free R-modules.

The complex is exact, since it is acyclic.

Since \(A_n\) is a free module, we can decompose its basis into a direct sum \(A_n = Z_n \oplus B'_n \) where \(Z_n\) is the kernel of the chain map \(d_n\) and \(B'_n = A_n \ Z_n\). Since the complex is exact, we have that \(Z_n = d(C_n) = B'_{n+1}\). So we can choose a basis in \(B'_{n+1}\) and let \(s_n:C_n \to C_{n+1}\) be s_n(z_i) = j_k where \(z_i\) is an element of the basis for \(Z_n\) and \(j_k\) is a fixed choice of basis element of \(B'_{n+1}\) and \(s_n(0) = 0\). Our choice of \(j_k\) is possible since free \(R\)-modules are projective. We can choose a basis for \(Z_n\) and let \(s_n: A_n \to A_{n+1}\) be the map which sets \(ds= id\) on \(Z_n\) via the projectivity of \(Z_n\). So, we have a map \(s\) such that \(dsd = d\). Hence, \(A.\) splits

Allowing us to conclude that \(A.\) is split exact.

Let \(A.\) be an acyclic chain complex of finitely generated free abelian groups.

The complex is exact, since it is acyclic.

Since \(A_n\) is a finitely generated free group, we can decompose it into a direct sum \(A_n = Z_n \oplus B'_n\) where \(Z_n\) is the kernel of the chain map \(d_n\) and \(B'_n = A_n \ Z_n\). Since the complex is exact, we have that \(Z_n = d(C_n) = B'_{n+1}\). So we can choose a basis in \(B'_{n+1}\) and let \(s_n:C_n \to C_{n+1}\) be s_n(e_i) = f_i where \(d(f_i) = e_i\) and \(e_i\) is an element of the basis for \(Z_n\) since the kernel is also free. \(f_i\) is an element of the basis of \(B'_{n+1}\) and \(s_n(0) = 0\). Our choice of \(f_i\) is possible since free abelian groups are projective. We can choose a basis for \(Z_n\) and let \(s_n: A_n \to A_{n+1}\) be the map which sets \(ds= id\) on \(Z_n\) via the projectivity of \(Z_n\). So, we have a map \(s\) such that \(dsd = d\). Hence, \(A.\) splits

Thus, \(A.\) is split exact.

Let cone\((C)\) denote the mapping cone of the identity map \(id_C\) of \(C\); it has \(C_{n-1} \oplus C_{n}\) in degree \(n\). Show that cone\((C)\) is split exact, with \(s(b,c)=(-c,0)\) defining the splitting map.

The mapping cone of the identity has \(d\) defined as \(d(a,b) = (-d(a),d(b) - a)\).

We check that \(dsd=d\):

\( \begin{split} d(s(d(a,b))) = d(s(-d(a),d(b) - a)) & = d(a-d(b), 0) \\ &= d(-d(a), 0 - (a-d(b))) \\ &= d(-d(a), d(b) -a) \\ &= d(a,b) \end{split}\)

Thus, \(dsd = d\) and \(cone(C)\) is split.

To see exactness, since the identity map is a quasi-isomorphism, the mapping cone of the identity is exact, by Corrolary 1.5.4. . What would a direct proof look like?