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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 $L=\{m_1(1,0)+m_2(0,1):m_1,m_2\in\mathbb{Z}$}

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    You 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

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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 $x=y$.

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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?
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    *this only works once you fix the typo @generic mentions.2012-07-24