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Homework answers / question archive / Let I be a non-empty index set with a partial order <=, and A_i be a group for all i in I
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
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.
Proof:
(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
phi_ji(a_j*b_j^(-1))=phi_ji(a_j)*phi_ji(b_j^(-1))
=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 groups A_i
(b) Since each phi_ji is surjective, then we consider an arbitrary element
a_i in A_i. For all i<=j, we can find some a_j in A_j, such that
phi_ji(a_j)=a_i. Then such x=(a_i) satisfies the definition of P and
thus belongs to P. So phi_i(x)=a_i. So phi_i is also surjective.
Since phi_i is surjective, according to the first group homomorphism
theorem, P/ker(phi_i) is isomorphic to A_i and thus A_i is a quotient
group of P.
(c) I think here A means P because you did not give definition of A.
So we give a ring structure of P like the followings.
For any (a_i), (b_i) in P, we have
(a_i)+(b_i)=(a_i+b_i), (a_i)*(b_i)=(a_i*b_i)
Now we show that this definition makes sense.
Because each phi_ji is a ring homomorphism for all i<=j, then
phi_ji(a_j+b_j)=phi_ji(a_j)+phi_ji(b_j)=a_i+b_i
phi_ji(a_j*b_j)=phi_ji(a_j)*phi_ji(b_j)=a_i*b_i
So the above (a_i+b_i) and (a_i*b_i) are all in P.
From (b), we know that phi_i is a ring homomorphism.
(d) First, I definite pi: K->P such that for any k in K, we have
pi(k)=(pi_i(k)). Since pi_i is a group homomorphism, then pi
is also a group homomorphism. We also have
phi_ipi(k)=phi_i((pi_i(k)))=pi_i(k) and thus get phi_ipi=pi_i.
The only thing we need to check is
that (pi_i(k)) is really in P.
For any i<=j, since pi_i=phi_jipi_j, then we have
phi_ji(pi_j(k))=pi_i(k) and this exactly satisfies the definition of P.
So pi: K->P is a well-defined group homomorphism.
Next we verify the uniqueness of pi. Suppose we also define pi':K->P
such that pi'(k)=(pi_i'(k)) for any k in K, then
phi_ipi'(k)=pi_i'(k)=pi_i(k) and thus pi'=pi. So the uniqueness is done.
