Ext and Balanced Resolutions — Examples

The augmentation identifies $M=\mathrm{coker}(d_1:P_1\to P_0)$, so left exactness of $\mathrm{Hom}(-,N)$ identifies $\ker (\mathrm{Hom}(P_0,N)\to \mathrm{Hom}(P_1,N))$ with $\mathrm{Hom}(M,N)$. Thus ${\mathrm{Ext}}_P^0(M,N)=\mathrm{Hom}(M,N)$.

Likewise $N=\ker (d^0:I^0\to I^1)$, and left exactness of $\mathrm{Hom}(M,-)$ gives $\ker (\mathrm{Hom}(M,I^0)\to \mathrm{Hom}(M,I^1))=\mathrm{Hom}(M,N)$. The two identifications are the degree-zero maps used in balanced ${\mathrm{Ext}}^0(M,N)$.

Applying $\mathrm{Hom}(-,N)$ gives the cochain complex $0\to \mathrm{Hom}(P_0,N)\stackrel{u^{\ast }}{\to }\mathrm{Hom}(P_1,N)\to 0$, where $u^{\ast }(f)=f\circ u$. Its degree-zero kernel is $\mathrm{Hom}(M,N)$.

Consequently ${\mathrm{Ext}}^1(M,N)=\mathrm{coker}{u}^{\ast }$, while ${\mathrm{Ext}}^q(M,N)=0$ for $q\ge 2$ because the displayed complex has no terms in those degrees. The projective-resolution computation is independent of this chosen two-term resolution.

The sequence $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\mathbb{Z}\to 0$ is a projective resolution: multiplication by $n$ is injective and its cokernel is $\mathbb{Z}\mathrm{/}n\mathbb{Z}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}}(-,A)$ gives $0\to A\stackrel{n}{\to }A\to 0$.

Therefore ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Z}\mathrm{/}n\mathbb{Z},A)=\mathrm{coker}(n:A\to A)=A\mathrm{/}nA$, and the same two-term complex has zero cohomology in every degree $q\ge 2$.

The low square has $K^{0,0}=\mathrm{Hom}(P_0,I^0)$, $K^{1,0}=\mathrm{Hom}(P_1,I^0)$, $K^{0,1}=\mathrm{Hom}(P_0,I^1)$, and $K^{1,1}=\mathrm{Hom}(P_1,I^1)$. Its unsigned arrows are $d_h(f)=f{d}_P$ and $d_v(f)=d_{I}f$, and $d_h{d}_v=d_v{d}_h$ on each $f$.

With $D{\mathrm{\mid }}_{K^{p,q}}=d_h+(-1{)}^p{d}_v$, ${\mathrm{Tot}}^0=K^{0,0}$ and ${\mathrm{Tot}}^1=K^{1,0}\oplus K^{0,1}$, so $D(f)=(f{d}_P,d_{I}f)$. For $(a,b)\in {\mathrm{Tot}}^1$, its $K^{1,1}$ component is $-d_{I}a+b{d}_P$, displaying the sign which makes $D^2=0$.

The first syzygy of $M$ for this presentation is $\mathrm{\Omega }M\cong \mathbb{Z}$, which is free and hence projective. The low-degree portion of the long exact sequence is $0\to \mathrm{Hom}(M,N)\to N\stackrel{n}{\to }N\to {\mathrm{Ext}}^1(M,N)\to 0$.

For every $q\ge 1$, dimension shifting gives ${\mathrm{Ext}}^{q+1}(M,N)\cong {\mathrm{Ext}}^q(\mathbb{Z},N)=0$. This does not replace the displayed low-degree segment: its cokernel is the generally nonzero group $N\mathrm{/}nN={\mathrm{Ext}}^1(M,N)$.

The group $\mathbb{Q}\mathrm{/}\mathbb{Z}$ is divisible, hence injective, so it is an injective object in the first variable. The displayed sequence is an extension of $\mathbb{Q}\mathrm{/}\mathbb{Z}$ by $\mathbb{Z}$.

This extension does not split: a retraction $r:\mathbb{Q}\to \mathbb{Z}$ would satisfy $r{\mathrm{\mid }}_{\mathbb{Z}}=1$, whereas every homomorphism $\mathbb{Q}\to \mathbb{Z}$ is zero (if $r(1)=a$, then $a=r(1\mathrm{/}m)m$ is divisible by every $m$). Hence it represents a nonzero element of ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q}\mathrm{/}\mathbb{Z},\mathbb{Z})$, refuting the asserted first-variable vanishing.

Precomposition by the projective comparison for $a$ and postcomposition by the injective comparison for $b$ define a morphism of first-quadrant Hom double complexes $\mathrm{Hom}(P(M),I(N))\to \mathrm{Hom}(P(M^{\mathrm{\prime }}),I(N^{\mathrm{\prime }}))$. It commutes with both $d_h$ and $d_v$.

The induced map on the total complex restricts on the projective edge and the injective edge to the usual maps on their Hom complexes. Passing to cohomology makes the two edge-to-total quasi-isomorphism squares commute; therefore the balance isomorphism intertwines ${\mathrm{Ext}}^q(M,N)\to {\mathrm{Ext}}^q(M^{\mathrm{\prime }},N^{\mathrm{\prime }})$ for every $q$.