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.

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 ...

Solution Summary

Direct Products, Mappings and Homomorphisms are investigated.

Let G = (Z/3Z)^4 SemiDirectProduct S_4 be the semi-direct product of (Z/3Z)^4 and S_4. Here S_4 acts on (Z/3Z)^4 by permutating the coordinates.
Hint: Given H1, H2 an element in (Z/3Z)^4 and K1, K2 an element in S4. The semi-direct product is given by the operation (H1, K1) * (H2, K2) = (H1 + K1(H2), K1 * K2)
A) Find the C

Topology
Sets and Functions (XXXIII)
Functions
Two mappings f : X --> [Y and g : X --> Y are said to be equal ( and we write this f = g )
if f(x) = g(x) for every x in X.

1. Let f : X -> Y and g : Y -> Z be mappings.
(1) Show that if f and g are both injective, then so is g o f : X -> Z
(2) Show that if f and g are both surjective, then so is g o f : X -> Z.
2. Let alpha = 1 2 3 4 5 and Beta = 1 2 3 4 5
3 5 1 2 4 3 2 4 5 1 .

Please help with the following problems. Provide step by step calculations for each.
1) Show that this mapping is linear:
T: P5 -> P8 defined as Tp(t)=p(t+1)-p(t)+integral(t-1 to t) s^2 p(s) ds
2) Prove the following is true, or give a counterexample:
If l is a nonzero scalar linear function on linear space X (which may

1) Prove that the map GIVEN BY is a homomorphism between the real line and open interval (-1,1).
2) Let be the map given by
a) show that f is a bijection map
b) show that f is a continuous map
c) If f a homomorphism? Justify your answer.
Please see the attached file for the fully formatted problems.

Let I be a non-empty index set with a partial order<=. Assume that I is a directed set, that is, that for any pair i,j in I there is a,k in J such that i<=k and j<=k. Suppose that for every pair of indices i,j with i<=j ther is a map p_ij: A_i->A_j such that p_jkp_ij=p_ik whenever i<=j<=k and p_ii=1 for all i in I. Let B be the

1)Let X={1,2,...,n}and let R be the Boolean ring of all subsets of X.
Define f_i:R->Z_2 by f_i(a)=[1] iff i is in a.Show each f_i is a
homomorphism and thus f=(f_1,...,f_n):R->Z_2*Z_2*...*Z_2 is a ring
homomorphism.Show f is an isomorphism.
2)If T is any ring,an element e of T is called an idempotent provided
e^2=e.The el