Derived Categories — Examples

The complex $U$ has $H^{-1}=0$ and $H^0=\mathbb{Z}\mathrm{/}2$, and $s$ induces the identity on this quotient. The map $f$ is a chain map because the target has only degree $-1$. Thus $s$ is a projective resolution of $\mathbb{Z}\mathrm{/}2[0]$, and [F1] sends the cocycle $f$ to exactly the displayed roof.

The free resolution $U$ computes Ext: the degree-one Hom quotient is $\mathbb{Z}\mathrm{/}2\mathbb{Z}$, and $f$ is the cocycle $1$. The lift used in [F2] for the displayed short exact sequence is the identity on the middle copy of $\mathbb{Z}$, so its terminal cocycle is this same $f$. The degree-one sign conversion in [F1] gives $-f$, but $f$ and $-f$ differ by the boundary $2f$ in this Hom quotient. Thus the displayed roof represents the extension class and is its nonzero generator. The extension cannot split because every homomorphism $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ is zero.

Reduction modulo two after multiplication by two is zero. The map two is monic, its image equals the kernel of reduction, and reduction is epic; all other terms are zero. Thus every cohomology group vanishes, and the acyclicity criterion gives $QX=0$.

If $X$ were zero in $K$, its identity would be nullhomotopic. The equation in degree two would give $1_{\mathbb{Z}\mathrm{/}2}=d^1{h}^2$. But $h^2:\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ is zero, so this is impossible. Hence $X$ is nonzero in $K$.

Multiplication by two on $\mathbb{Z}$ is injective with cokernel $\mathbb{Z}\mathrm{/}2$. Thus $s$ is a quasi-isomorphism in all degrees, including the two boundary degrees and the zero terms elsewhere.

The inversion rule for localization identifies the displayed reversed roof with $Q(1_U)Q(s{)}^{-1}$. Composing on either side with $Q(s)$ gives the corresponding identity, proving the asserted inverse explicitly.

Resolve $\mathbb{Z}\mathrm{/}m$ by $P=(\mathbb{Z}\stackrel{m}{\to }\mathbb{Z})$ in degrees $-1,0$, with its quotient augmentation. Since $m>0$ the first map is injective, so this is a projective resolution. Hom into $\mathbb{Z}\mathrm{/}n$ has terms $\mathbb{Z}\mathrm{/}n$ in degrees zero and one, and differential $-m$ with the cochain Hom convention.

Degree-one cohomology is $(\mathbb{Z}\mathrm{/}n)\mathrm{/}m(\mathbb{Z}\mathrm{/}n)=\mathbb{Z}\mathrm{/}(m\mathbb{Z}+n\mathbb{Z})=\mathbb{Z}\mathrm{/}\gcd (m,n)$. The Ext comparison identifies it with the claimed Hom group. If $m=1$ or $n=1$ the quotient is zero, as required for a zero input module.

Upper brutal truncation retains the degree-zero $\mathbb{Z}$ and deletes the next term. Upper canonical truncation replaces it by $\ker (2)=0$. All lower degrees are already zero.

The lower brutal truncation deletes only already-zero terms. The lower canonical boundary is $\mathrm{coker}(0\to \mathbb{Z})=\mathbb{Z}$, so its differential remains multiplication by two. Thus both lower truncations equal $X$. Its $H^0$ is zero, agreeing with the canonical upper truncation and disagreeing with the brutal upper truncation. The four constructions need not be four distinct complexes.

The kernel and cokernel of multiplication by two are $2R$ and $R\mathrm{/}2R$. Set $A=(2R)[0]$. The quotient complex $V=X\mathrm{/}A$ has $V^0=R\mathrm{/}2R,V^1=R$ with differential $\bar{r}\mapsto 2r$. The map $v:V\to (R\mathrm{/}2R)[-1]$ is quotient in degree one and zero in degree zero. Its kernel is the identity complex on $2R$ after identifying the degree-zero term with $2R$, so $v$ is a quasi-isomorphism.

Let $C=\mathrm{Cone}(i)$, with $C^k=X^k\oplus A^{k+1}$ and differential $(x,a)\mapsto (d_{X}x+i(a),-d_{A}a)$. The cone-to-quotient map $e:C\to V$ sends $(x,a)$ to the class of $x$; its kernel is the contractible identity cone on $A$. Thus $e$ is a quasi-isomorphism. Let $p:C\to A[1]$ be $(x,a)\mapsto a$. Then $\delta$ is the roof $(R\mathrm{/}2R)[-1]\stackrel{ve}{\leftarrow }C\stackrel{p}{\to }A[1]$. The cone triangle transported through $ve$ has first arrow $i$ and second arrow $q=v(X\to V)$, proving every displayed arrow with the stated signs.

Use the free resolution $(\mathbb{Z}\stackrel{m}{\to }\mathbb{Z})\to \mathbb{Z}\mathrm{/}m$ and tensor with $\mathbb{Z}\mathrm{/}n$. This gives exactly the displayed two-term complex; since $m>0$ the resolution is exact at its left endpoint. Its cohomology is its kernel at $-1$ and cokernel at zero.

Put $g=\gcd (m,n)$. The cokernel is $\mathbb{Z}\mathrm{/}(m\mathbb{Z}+n\mathbb{Z})=\mathbb{Z}\mathrm{/}g$. The kernel consists of the multiples of $n\mathrm{/}g$ modulo $n$, and $\mathbb{Z}\mathrm{/}g\to \ker (m)$, $\bar{k}\mapsto \overline{kn\mathrm{/}g}$, is an isomorphism: $n\mid mk$ iff $n\mathrm{/}g\mid k$. The kernel in degree $-1$ is ${\mathrm{Tor}}_1$, and the cokernel in degree zero is ${\mathrm{Tor}}_0$. If either modulus is one both groups are zero; all other degrees have zero terms.

Apply the Hom complex to the free resolution $(\mathbb{Z}\stackrel{m}{\to }\mathbb{Z})$ in degrees $-1,0$ and the target $\mathbb{Z}\mathrm{/}n[0]$. Its two terms are $\mathbb{Z}\mathrm{/}n$ in degrees zero and one with differential $-m$. Multiplication by $-1$ in degree one and identity in degree zero gives a complex isomorphism to the displayed $+m$ model.

For $g=\gcd (m,n)$, the kernel of $m$ is generated by $n\mathrm{/}g$ modulo $n$ and is cyclic of order $g$; the cokernel is $\mathbb{Z}\mathrm{/}(m\mathbb{Z}+n\mathbb{Z})=\mathbb{Z}\mathrm{/}g$. These are respectively $\mathrm{Hom}(\mathbb{Z}\mathrm{/}m,\mathbb{Z}\mathrm{/}n)$ and ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}m,\mathbb{Z}\mathrm{/}n)$ by derived Hom cohomology. Modulus one makes both zero, and all other degrees vanish.

Set $R=\mathbb{Z}\mathrm{/}4$, and take $P^i=R$ and $d^i=2$ for every integer $i$. The terms are free of rank one and thus projective (a map out of $R$ lifts by lifting its value at $1$). Also $d^{i+1}{d}^i=4=0$, and $\ker d^i=2R=\mathrm{im}{d}^{i-1}$. Thus $P$ is an acyclic doubly infinite complex.

Every proposed homotopy $h^i:R\to R$ is multiplication by an element $a_i$. The identity-homotopy equation at degree $i$ would require $1=2a_i+2a_{i+1}$, impossible modulo two. Hence $1_P$ is nonzero in ${\mathrm{Hom}}_K(P,P)$. Since $P$ itself is acyclic, this contradicts the vanishing required for a K-projective source. The obstruction is an equation at every degree, not evidence from a finite truncation.