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Direct Products, Mappings and Homomorphisms

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Let I be a non-empty index set with a partial order <=, and A_i be a group for all i in I. Suppose that for every pair of indices i,j there is a map phi_ij:A_j ->A_i such that phi_jiphi_kj= phi_ki wheneveri<=j<=k, and phi_ii=1 for all i in I. Let P be the subset of elements(a_i) with i from I in the direct product D of A_i such that phi_ij(a_j)=a_i whenever i<=j. P is called the inverser or projective limit of the system {A_i} and is denoted lim<-A_i.
a) Assume all phi_ji are group homomorphisms. Show that P is a subgroup of the direct product of the groups A_i.
b) Assume I=Z+ and phi_ji are group homomorphisms. For each i in I let phi_i:P->A_i be the projection of P onto its i-th component.Show that if each phi_ji is surjective, then so is phi_i, for all i so that each A_i is a quotient group of P
c)Show that if all A_i are commutative rings with 1 and all phi_ji are ring homomorphisms that send 1 to 1, then A may likewise be given the structure of a commutative ring with 1 such that all phi_i are ring homomorphisms.
d)Assume all phi_ji are group homomorphisms. Prove that the inverse limit has the following universal mapping property: if K is any group such that for each i in I there is a homomorphism pi_i:K->A_i with pi_i=phi_jipi_j whenever i<=j, then there is a unique homomorphism pi: K->P such that phi_ipi=pi_i for all i in I

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I think the definition of phi_ij is from A_i to A_j, not from A_j to A_i, so that
phi_ki=phi_ji phi_kj makes sense. So (a_i) in P means phi_ji(a_j)=a_i for all i<=j.

(a) Consider any two elements (a_i) and (b_i) in P. The inverse of (b_i) is (b_i^(-1)).
Then we want to show that (a_i)*(b_i)^(-1)=(a_i*b_i^(-1)) is in P.
Because all phi_ji are group homomorphisms, then for any i<=j, we have
=a_i*b_i^(-1). Then (a_i*b_i^(-1)) is in P and thus P is a subgroup of the direct
product of the ...

Solution Summary

Direct Products, Mappings and Homomorphisms are investigated.