Yoneda Extensions and Homological Dimension — Examples

The pullback of $e\oplus s$ along $\mathrm{\Delta }:M\to M\oplus M$ is canonically $E\oplus N$: the morphism $(1_E,1_N\oplus p):E\oplus N\to E\oplus (N\oplus M)$ exhibits the pullback. Under this identification its kernel map is $(a,b)\mapsto (i(a),b)$ from $N\oplus N$.

The morphism $(1_E,i):E\oplus N\to E$ agrees on that kernel with $i\circ \mathrm{\nabla }:N\oplus N\to E$. Hence the universal property of the pushout along $\mathrm{\nabla }$ gives a morphism from the Baer-sum middle object to $E$ that is the identity on both endpoints. A morphism of short exact sequences that is the identity on the endpoints is an isomorphism, so the sum is equivalent to $e$. Thus the split class is zero without using elements or module quotients.

Any middle group in an extension of $\mathbb{Z}\mathrm{/}2$ by $\mathbb{Z}\mathrm{/}2$ has four elements. Transporting its law and endpoint maps to a fixed four-element set shows that the classes have a finite set realization. The extension $e$ is nonsplit because every element of $\mathbb{Z}\mathrm{/}4$ above $1\in \mathbb{Z}\mathrm{/}2$ has order four.

The diagonal pullback has middle group $D=\{(x,y)\in (\mathbb{Z}\mathrm{/}4{)}^2:x\equiv y(\mathrm{m}\mathrm{o}\mathrm{d}2)\}$ and kernel map $(a,b)\mapsto (2a,2b)$. Its codiagonal pushout is $E=(D\oplus \mathbb{Z}\mathrm{/}2)\mathrm{/}L$, where $L=\{((2a,2b),-(a+b)):a,b\in \mathbb{Z}\mathrm{/}2\}$. The endpoint maps are $j(c)=[((0,0),c)]$ and $q([((x,y),c)])=x<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$.

The map $j$ is injective since a relation with first component $(0,0)$ has $a=b=0$. If $q([((x,y),c)])=0$, write $x=2a,y=2b$; its class equals $j(c+a+b)$. Thus $\ker q=\mathrm{im}j$, and $q$ is surjective. The class $t=[((1,-1),0)]$ has $q(t)=1$ and $2t=[((2,2),0)]=0$, the relation for $a=b=1$.

Therefore $s(1)=t$ defines a homomorphic section of $q$. The map $(c,b)\mapsto j(c)+s(b)$ is an endpoint-preserving isomorphism $(\mathbb{Z}\mathrm{/}2{)}^2\to E$: surjectivity and injectivity follow from the kernel description and $qs=1$. Hence the Baer sum is split and equals the zero class.

Put $E_a=(A\oplus \mathbb{Z})\mathrm{/}⟨(a,-n)⟩$. The maps $A\to E_a$, $x\mapsto [(x,0)]$, and $E_a\to \mathbb{Z}\mathrm{/}n\mathbb{Z}$, $[(x,k)]\mapsto [k]$, give an exact sequence $0\to A\to E_a\to \mathbb{Z}\mathrm{/}n\mathbb{Z}\to 0$.

Replacing $a$ by $a+nb$ gives an equivalent extension by changing the lift of $1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>n$ by the image of $b\in A$. Conversely the connecting class is the residue of $a$ modulo $nA$. Thus the derived/Yoneda correspondence identifies $A\mathrm{/}nA$ with these extension classes.

Concatenating the maps gives $0\to N\to E\stackrel{vu}{\to }F\to M\to 0$. Its exactness at $E$ follows from $v$ monic and $\ker u=N$. At $F$, $\mathrm{im}(vu)=v(\mathrm{im}u)=v(L)$, which is exactly the kernel of $F\to M$.

This exact five-term sequence is the two-fold Yoneda extension obtained by splicing. Under the comparison with derived Ext, its class is the composition of the classes of the two displayed short exact sequences in ${\mathrm{Ext}}^2(M,N)$.

Let $V=(x,y)$, so $V^2=0$. The free resolution $P_n=R{\otimes }_k{V}^{\otimes n}$ with differential given by multiplication by the first $V$-factor has $\ker d_n=V{\otimes }_k{V}^{\otimes n}=\mathrm{im}{d}_{n+1}$. Applying ${\mathrm{Hom}}_R(-,S)$ gives zero differentials, so ${\mathrm{Ext}}_R^n(S,S)\cong (V^{\ast }{)}^{\otimes n}$. Let $u,v$ be the dual basis classes in degree one.

Yoneda splicing corresponds to concatenating tensors under this resolution. Consequently $uv$ and $vu$ are distinct basis vectors of $(V^{\ast }{)}^{\otimes 2}={\mathrm{Ext}}_R^2(S,S)$. Both products are typed in the same self-Ext algebra, and they differ; this is a genuinely noncommutative Yoneda product.

The exact sequence $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to M\to 0$ is a projective resolution of length one, so ${\mathrm{pd}}_{\mathbb{Z}}M\le 1$.

Applying ${\mathrm{Hom}}_{\mathbb{Z}}(-,\mathbb{Z})$ gives ${\mathrm{Ext}}_{\mathbb{Z}}^1(M,\mathbb{Z})\cong \mathbb{Z}\mathrm{/}n\mathbb{Z}\mathrm{\ne }0$. The higher-Ext criterion therefore rules out projective dimension zero, so ${\mathrm{pd}}_{\mathbb{Z}}M=1$.

Every $k$-module is a vector space and has a basis, hence is free and projective. Therefore every module has projective dimension zero and $\mathrm{gl.dim}k=0$.

For every abelian group $A$, a free presentation has free kernel, so ${\mathrm{pd}}_{\mathbb{Z}}A\le 1$ by The integers have global dimension one. The concrete nonzero group ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Z}\mathrm{/}2,\mathbb{Z})\cong \mathbb{Z}\mathrm{/}2$ supplies a degree-one witness. Hence $\mathrm{gl.dim}\mathbb{Z}=1$.

Let $E$ be $0\to \mathbb{Z}\stackrel{1}{\to }\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\stackrel{1}{\to }\mathbb{Z}\to 0$, and let $E^{\mathrm{\prime }}$ be $0\to \mathbb{Z}\to {\mathbb{Z}}^2\stackrel{(a,c)\mapsto (0,c)}{\to }{\mathbb{Z}}^2\stackrel{(b,c)\mapsto b}{\to }\mathbb{Z}\to 0$, where the left inclusion is $a\mapsto (a,0)$. Both are exact.

The maps $(a)\mapsto (a,0)$ and $(b)\mapsto (b,0)$ give an endpoint-identity chain map $E\to E^{\mathrm{\prime }}$, while the coordinate projections give one $E^{\mathrm{\prime }}\to E$. Thus the two extensions are equivalent in the generated sense. But their first middle terms are $\mathbb{Z}$ and ${\mathbb{Z}}^2$, which are not isomorphic, so no single middle-term isomorphism can witness this equivalence.