What goes wrong when arrows are reversed in the diagrams for direct product/sum universal properties?

I’m taking an algebra course and regaining my algebra sense (non-sense?). Familiar to us is that direct products [resp. sums] of groups [resp. abelian groups] possess their versions of universal properties. When reviewing these things, I went curious on what would go wrong if the arrows were reversed. Let’s try and see.

Correct Version of Universal Properties

Direct Sum of Abelian Groups

Let (Ai)iI{\left(A_i\right)}_\idxi be a non-empty family of abelian groups and A=iIAiA=\bigoplus_\idxi{A_i} their direct sum. Let ηi:AiA\eta_i:A_i\to A be canonical injections for all iI\idxi. For all abelian group BB and a family of homomorphisms φi:AiB,iI\varphi_i:A_i\to B,\forall\idxi, there exists a unique homomorphism φ:AB\varphi:A\to B such that φηi=φi\varphi\eta_i=\varphi_i for all iI\idxi, i.e., the following diagrams commute for all iI\idxi: φ η i equals φ i for all i in I.φ η i equals φ i for all i in I.φ η i equals φ i for all i in I.φ η i equals φ i for all i in I.φ η i equals φ i for all i in I. This homomorphism is φ((ai)iI)=iIxφi(ai)\displaystyle\varphi\left({\left(a_i\right)}_\idxi\right)=\sum_\idxi^x{\varphi_i\left(a_i\right)}.

Direct Product of Groups

Let (Gi)iI{\left(G_i\right)}_\idxi be a non-empty family of groups and G=iIGiG=\prod_\idxi{G_i} their direct product. Let πi:GGi\pi_i:G\to G_i be canonical projections for all iI\idxi. For all group HH and a family of homomorphisms φi:HGi\varphi_i:H\to G_i, there exists a unique homomorphism φ:HG\varphi:H\to G such that πiφ=φi\pi_i\varphi=\varphi_i for all iI\idxi, i.e., the following diagrams commute for all iI\idxi: π i φ equals φ i for all i in I.π i φ equals φ i for all i in I.π i φ equals φ i for all i in I.π i φ equals φ i for all i in I.π i φ equals φ i for all i in I. This homomorphism is φ(h)=(φi(h))iI\displaystyle\varphi\left(h\right)={\left(\varphi_i\left(h\right)\right)}_\idxi.

Incorrect Version of Universal Properties

We now explore what will go wrong if the arrows are reversed in the above diagrams. All symbols not redefined (or requantified) are assumed to have the same meaning as defined above.

⊕ for Π Properties

We still confine ourselves to abelian groups. Reversing the arrows means that for any abelian group BB and a family of homomorphisms φi:BAi\varphi_i:B\to A_i, we can induce a unique homomorphism φ:BA\varphi:B\to A such that πiφ=φi\pi_i\varphi=\varphi_i. The homomorphism in imagination might not exist.

For all aAa\in A, there exists a finite subset JI\mathcal{J}\subseteq\mathcal{I} such that πi(a)=0\pi_i\left(a\right)=0 for all iIJi\in\mathcal{I}\setminus\mathcal{J}. Since the image of the induced homomorphism must be in the direct sum, this puts a restriction on what family of φi\varphi_i one can choose. For example, consider I=N\mathcal{I}=\mathbb{N}, B=Ai=ZB=A_i=\mathbb{Z} and φi=idZ\varphi_i=\id_\mathbb{Z}. The image of 11 under the imaginary induced φ\varphi would not lie in the direct sum.

If such a homomorphism can be induced, it will be unique. Such a homomorphism exists if and only if the induced one for direct product maps only into the direct sum, and in that case, the induced homomorphism is simply the one for the direct product.

Π for ⊕ Properties


Under general group setting, the homomorphism in question might not exist. The factor groups are naturally embedded as disjoint normal subgroups of the product (i.e., the internal direct product decomposition). We use that to find a counterexample.

Let I={1,2}\mathcal{I}=\left\{{1,2}\right\}, G1=G2=HG_1=G_2=H and φi=idH\varphi_i=\id_H. If an induced homomorphism φ\varphi exists, we know that for all x,yHx,y\in H, xy=φ1(x)φ2(y)=φ(x,1)φ(1,y)=φ(x,y)=φ(1,y)φ(x,1)=φ2(y)φ1(x)=yx,\ProdSumFail which means HH must be abelian. Picking a non-abelian HH yields a counterexample.

Existence (Abelian)

The homomorphism in question might not exist even if we are working exclusively with abelian groups. For abelian groups, one can always induce the homomorphism from the direct sum to the codomain, which can be any possible such homomorphism. The question is thus equivalent to wheher arbitrary homomorphism from the direct sum to some codomain can be extended to one from the direct product. This turns out to be equivalent to whether the direct sum is a direct summand of the direct product (see remarks).

We have chosen the identity map for the previous two counterexamples. This time, it involves using the direct sum itself as the codomain. Let I=N\mathcal{I}=\mathbb{N}, Gi=ZG_i=\mathbb{Z}, H=iIGi\displaystyle H=\bigoplus_\idxi{G_i}, K=iIGi\displaystyle K=\prod_\idxi{G_i} and φi=ηi\varphi_i=\eta_i. Does there exists some φ\varphi such that the following diagrams commute? φ η i equals η i for all i in I.φ η i equals η i for all i in I.φ η i equals η i for all i in I.φ η i equals η i for all i in I.φ η i equals η i for all i in I. The answer is no. We prove by contradiction. Assume yes instead, by the universal property of direct sum of abelian groups, we have φH=idH{\left.\varphi\right|}_H=\id_H. This means xφ(x)=0x-\varphi\left(x\right)=0 for all xHx\in H. Now define x0=(1,2,4,8,16,),x1=(0,1,2,4,8,),x2=(0,0,1,2,4,),x3=(0,0,0,1,2,),yn=xφ(xn).\MultipleSeq Since x0Hx_0\notin H and φ(x0)H\varphi\left(x_0\right)\in H, we have y00y_0\neq 0. Moreover, x02nxnHx_0-2^nx_n\in H, therefore y0=2nyny_0=2^ny_n. This contradicts the fact that y0y_0 has a non-zero integral coordinate.

Professor Goodearl suggests another counterexample. Consider A=p primeZp\displaystyle A=\prod_{p\text{ prime}}{\mathbb{Z}_p}, then T(A)=p primeZp\displaystyle T\left(A\right)=\bigoplus_{p\text{ prime}}{\mathbb{Z}_p} and T(A)T\left(A\right) is not a direct summand of AA. (Again see remarks for equivalence of direct summand and homomorphism extension.)

Uniqueness (Abelian)

Suppose that there exists an induced homomorphism for the reversed version of universal property for direct sum of abelian groups. How do we know whether the induction yields a unique homomorphism? If not, how can we describe the possible inductions?

For abelian groups, Professor Goodearl points out that one can study the difference between two induced homomorphisms.

Given an abelian group homomorphism φ:iIAiB\displaystyle\varphi:\bigoplus_\idxi{A_i}\to B, consider two homomorphisms ψ1,ψ2\psi_1,\psi_2 that make the following diagram commute: ψ 1 ι and ψ 2 ι both equal φ.ψ 1 ι and ψ 2 ι both equal φ.ψ 1 ι and ψ 2 ι both equal φ.ψ 1 ι and ψ 2 ι both equal φ.ψ 1 ι and ψ 2 ι both equal φ. Define ν:iIAiB,aψ1(a)ψ2(a),\nu:\prod_\idxi{A_i}\to B,a\mapsto\psi_1\left(a\right)-\psi_2\left(a\right), and we can find that iIAikerν\displaystyle\bigoplus_\idxi{A_i}\leq\ker\nu. Moreover, for all homomorphism μ:iIAiB\displaystyle\mu:\prod_\idxi{A_i}\to B such that iIAikerμ\displaystyle\bigoplus_\idxi{A_i}\leq\ker\mu, we have (ψ1+μ)ι=φ\left(\psi_1+\mu\right)\iota=\varphi.

Therefore, the ‘solutions’ of the above commutative diagram are exactly those can be written as ψ+ξπ\psi^\star+\xi\pi, where ψ\psi^\star is one specific solution, ξHom(iIAiiIAi,B)\displaystyle\xi\in\Hom\left({\frac{\prod_\idxi{A_i}}{\bigoplus_\idxi{A_i}},B}\right), and π:iIAiiIAiiIAi\displaystyle\pi:\prod_\idxi{A_i}\to\frac{\prod_\idxi{A_i}}{\bigoplus_\idxi{A_i}} is the canonical map.

This resembles the structure of solutions to (not necessaily homogeneous) linear equations very much.

Remarks and Acknowledgements

Extension of Homomorphism (Abelian) ⟺ Direct Summand

Let AA be an abelian group and BAB\leq A a subgroup. TFAE:

  1. For all group GG and homomorphism φ:BG\varphi:B\to G, there exists a homomorphism φ:AG\varphi^\star:A\to G such that φB=φ{\left.\varphi^\star\right|}_B=\varphi.
  2. For all abelian group CC … (1 restricted to abelian groups)
  3. There exists a homomorphism φ:AB\varphi^\star:A\to B such that φB=idB{\left.\varphi^\star\right|}_B=\id_B.
  4. There exists a subgroup CAC\leq A such that A=BCA=B\oplus C.

1 ⟹ 2, 2 ⟹ 3 and 4 ⟹ 1 are trivial.

(3 ⟹ 4) Define ψ:AA,aaφ(a)\psi:A\to A,a\mapsto a-\varphi^\star\left(a\right) and let C=imψC=\im\psi. For all aAa\in A, a=φ(a)+ψ(a)a=\varphi^\star\left(a\right)+\psi\left(a\right), therefore A=B+CA=B+C. For all aBCa\in B\cap C, we have a=ψ(b)a=\psi\left(b\right) for some bAb\in A, and a=idB(a)=φ(a)=φ(ψ(b))=φ(bφ(b))=φ(b)φ(φ(b))=φ(b)idB(φ(b))=φ(b)φ(b)=0.\IntersectIsZero Therefore, A=BCA=B\oplus C.

The technique to construct the complement subgroup resembles that to construct a complement vector subspace from idempotent linear transformations (projections) very much.


I would like to thank Professor Goodearl and Shan Zhou for their discussions on this problem with me.

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