please follow this map from $\mathbb{C}$ to $\mathbb{C}/L$ is open map? ,let w be a non zero element of the lattice L so that |w|>2ϵ, fix such ϵ>0 and any $z_0$∈C and take an open disk of radius ϵ with centre at $z_0$, could you please tell me why $π:D\rightarrow π(D)$ is injective? what does it mean by a "lattice" of small disk in $\mathbb{C}$? what is π(D) pictorically? let my $L=\{m_1(1,0)+m_2(0,1):m_1,m_2\in\mathbb{Z}$}
map from $D$ to $\pi(D)$ is injective?
-
0You know, considering your username is **Patience**, you might have taken more than a couple of hours to digest the answers you received to your previous question. – 2012-07-24
2 Answers
The statement you gave cannot be correct because there are points with arbitrarily high magnitude in $L$, so that would imply that $\pi$ is injective over $\mathbb C$.
You need $|w|>2\varepsilon$ for every non-zero $w\in L$, not just a single one.
Once you make that correction, $\pi(x)=\pi(y) \iff y-x\in L$, but since $|y-x|\le 2\varepsilon$ this can only be true if $x=y$.
So interestingly enough, the answer is similar to the other one I gave here in concept. The important thing is understanding equivalence classes!
You have a map from a set to a set which is a quotient map. Elements of $\mathbb{C}/L$ are equivalence classes of $\mathbb{C}$ mod an equivalence relation. Hence two points in $\mathbb{C}$ will map to the same class if they are representatives of the same class. You need to show that that can't happen.
You should work out the following
- what is the equivalence relation in question?
- when are two points in $\mathbb{C}$ representatives of the same class?
- why cant two points in the neighborhood you describe be representatives of the same class, based on the first two things?
- this shows the map is injective. Why?
-
0*this only works once you fix the typo @generic mentions. – 2012-07-24