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Gerry Myerson
Gerry Myerson
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The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have $\Lambda_1,\Lambda_2\subseteq X$ and then we have   $\Lambda_1,\Lambda_2\subseteq \Lambda_1+\Lambda_2\subseteq X$. This means that we have quotient maps 
$X/\Lambda_1\rightarrow X/(\Lambda_1+\Lambda_2)$ and $X/\Lambda_2\rightarrow X/(\Lambda_1+\Lambda_2)$.$$X/\Lambda_1\rightarrow X/(\Lambda_1+\Lambda_2){\rm\quad and\quad }X/\Lambda_2\rightarrow X/(\Lambda_1+\Lambda_2).$$ Condition 2) then means that the two composites $\mathbb Q^n/\mathbb Z^n\rightarrow X/\Lambda_i \rightarrow X/(\Lambda_1+\Lambda_2)$ are equal.

As for your interpretation of condition 2) it seems to be more or less OK, though the way you have phrased it $\phi$ could very well be $0$. To understand what 2) means in concrete terms it is convenient to use some more advanced notions (which may or may not be unfamiliar to you). First the source and target of $\phi$ are torsion groups. We then have that for each prime $p$ the $p$-torsion of $\mathbb Q^n/\mathbb Z^n$ is taken into the $p$-torsion of $\mathbb Q\Lambda/\Lambda$. Any map of these $p$-torsion groups corresponds exactly to a $\mathbb Z_p$-module map (where $\mathbb Z_p$ is the ring of $p$-adic integers) $\phi_p\colon\mathbb Z_p^n\rightarrow \Lambda\bigotimes \mathbb Z_p$. This is very analoguous to the situation when we instead would start with a continuous map $\phi\colon\mathbb R^n/\mathbb Z^n\rightarrow \mathbb R\Lambda/L$ which would correspond to a map   $\phi_{\mathbb Z}\colon\mathbb Z^n\rightarrow\Lambda$.

The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have $\Lambda_1,\Lambda_2\subseteq X$ and then we have $\Lambda_1,\Lambda_2\subseteq \Lambda_1+\Lambda_2\subseteq X$. This means that we have quotient maps $X/\Lambda_1\rightarrow X/(\Lambda_1+\Lambda_2)$ and $X/\Lambda_2\rightarrow X/(\Lambda_1+\Lambda_2)$. Condition 2) then means that the two composites $\mathbb Q^n/\mathbb Z^n\rightarrow X/\Lambda_i \rightarrow X/(\Lambda_1+\Lambda_2)$ are equal.

As for your interpretation of condition 2) it seems to be more or less OK, though the way you have phrased it $\phi$ could very well be $0$. To understand what 2) means in concrete terms it is convenient to use some more advanced notions (which may or may not be unfamiliar to you). First the source and target of $\phi$ are torsion groups. We then have that for each prime $p$ the $p$-torsion of $\mathbb Q^n/\mathbb Z^n$ is taken into the $p$-torsion of $\mathbb Q\Lambda/\Lambda$. Any map of these $p$-torsion groups corresponds exactly to a $\mathbb Z_p$-module map (where $\mathbb Z_p$ is the ring of $p$-adic integers) $\phi_p\colon\mathbb Z_p^n\rightarrow \Lambda\bigotimes \mathbb Z_p$. This is very analoguous to the situation when we instead would start with a continuous map $\phi\colon\mathbb R^n/\mathbb Z^n\rightarrow \mathbb R\Lambda/L$ which would correspond to a map $\phi_{\mathbb Z}\colon\mathbb Z^n\rightarrow\Lambda$.

The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have $\Lambda_1,\Lambda_2\subseteq X$ and then we have   $\Lambda_1,\Lambda_2\subseteq \Lambda_1+\Lambda_2\subseteq X$. This means that we have quotient maps 
$$X/\Lambda_1\rightarrow X/(\Lambda_1+\Lambda_2){\rm\quad and\quad }X/\Lambda_2\rightarrow X/(\Lambda_1+\Lambda_2).$$ Condition 2) then means that the two composites $\mathbb Q^n/\mathbb Z^n\rightarrow X/\Lambda_i \rightarrow X/(\Lambda_1+\Lambda_2)$ are equal.

As for your interpretation of condition 2) it seems to be more or less OK, though the way you have phrased it $\phi$ could very well be $0$. To understand what 2) means in concrete terms it is convenient to use some more advanced notions (which may or may not be unfamiliar to you). First the source and target of $\phi$ are torsion groups. We then have that for each prime $p$ the $p$-torsion of $\mathbb Q^n/\mathbb Z^n$ is taken into the $p$-torsion of $\mathbb Q\Lambda/\Lambda$. Any map of these $p$-torsion groups corresponds exactly to a $\mathbb Z_p$-module map (where $\mathbb Z_p$ is the ring of $p$-adic integers) $\phi_p\colon\mathbb Z_p^n\rightarrow \Lambda\bigotimes \mathbb Z_p$. This is very analoguous to the situation when we instead would start with a continuous map $\phi\colon\mathbb R^n/\mathbb Z^n\rightarrow \mathbb R\Lambda/L$ which would correspond to a map   $\phi_{\mathbb Z}\colon\mathbb Z^n\rightarrow\Lambda$.

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Torsten Ekedahl
Torsten Ekedahl
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The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have $\Lambda_1,\Lambda_2\subseteq X$ and then we have $\Lambda_1,\Lambda_2\subseteq \Lambda_1+\Lambda_2\subseteq X$. This means that we have quotient maps $X/\Lambda_1\rightarrow X/(\Lambda_1+\Lambda_2)$ and $X/\Lambda_2\rightarrow X/(\Lambda_1+\Lambda_2)$. Condition 2) then means that the two composites $\mathbb Q^n/\mathbb Z^n\rightarrow X/\Lambda_i \rightarrow X/(\Lambda_1+\Lambda_2)$ are equal.

As for your interpretation of condition 2) it seems to be more or less OK, though the way you have phrased it $\phi$ could very well be $0$. To understand what 2) means in concrete terms it is convenient to use some more advanced notions (which may or may not be unfamiliar to you). First the source and target of $\phi$ are torsion groups. We then have that for each prime $p$ the $p$-torsion of $\mathbb Q^n/\mathbb Z^n$ is taken into the $p$-torsion of $\mathbb Q\Lambda/\Lambda$. Any map of these $p$-torsion groups corresponds exactly to a $\mathbb Z_p$-module map (where $\mathbb Z_p$ is the ring of $p$-adic integers) $\phi_p\colon\mathbb Z_p^n\rightarrow \Lambda\bigotimes \mathbb Z_p$. This is very analoguous to the situation when we instead would start with a continuous map $\phi\colon\mathbb R^n/\mathbb Z^n\rightarrow \mathbb R\Lambda/L$ which would correspond to a map $\phi_{\mathbb Z}\colon\mathbb Z^n\rightarrow\Lambda$.