Is the Operation Linear and Bijective?

schw
Messages
3
Reaction score
0
Could anyone help me solve this problem?

Let A,B be two subspace of V, a \in A, b \in B. Show that the following operation is linear and bijective:

(A + B)/B → A/(A \cap B): a + b + B → a + A \cap B

I really couldn't understand how the oparation itself works, i.e, what F(v) really is in this problem.
 
Physics news on Phys.org
write v as a+b, then F(v) = a. now chck this is well defined, i.e. if a+b ≈ a'+b' modb, then a ≈ a' mod (AmeetB).
 
I just have to ignore the sets that are in the operation? (v = a + b + B)

edited:
Just realized the definition of quocient space. Tahnks for helping
 
Last edited:
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top