Let $\Lambda_1, \Lambda_2$ be lattices of $\mathbb{C}$ and $T:\mathbb{C}\rightarrow\mathbb{C}$ be a $\mathbb{C}$-linear map such that $T(\Lambda_1) \subset \Lambda_2$. This induces complex tori morphism $\varphi:\mathbb{C}/\Lambda_1 \rightarrow \mathbb{C}/\Lambda_2$.
A book I'm reading right now asserts that $\operatorname{Ker}(\varphi) \cong \Lambda_2/T(\Lambda_1)$.
Is this true? This seemed elementary so I glossed over it initially but now I'm having trouble showing this.
To me, $\operatorname{Ker}(\varphi)$ are the elements of $\mathbb{C}$ sent to $\Lambda_2$ under $T$, elements which are then considered mod $\Lambda_1$. Therefore, we have $\operatorname{Ker}(\varphi) \cong T^{-1}(\Lambda_2)/\Lambda_1$, with a well-defined quotient since $T$ sends $\Lambda_1$ into $\Lambda_2$.
Then I would be tempted to apply the map induced by $T$ to get $T^{-1}(\Lambda_2)/\Lambda_1 \xrightarrow{T} \Lambda_2/T(\Lambda_1)$.
But is this an isomorphism? I have doubts on this, it doesn't seem to be injective or surjective at first sight. For instance nothing tells me that if $x \in T^{-1}(\Lambda_2) \setminus\Lambda_1$, it won't land under $T$ in $T(\Lambda_1)$ (unless if $T^{-1}(T(\Lambda_1))=\Lambda_1$ but this is not clear to me why this would be so), which would contradict injectivity... I only know it lands in $\Lambda_2$ but $\Lambda_2$ contains $T(\Lambda_1)$ so this is not obvious...
Nicely framed question. Indeed $$ T^{-1}(\Lambda_2)/\Lambda_1 \xrightarrow{T} \Lambda_2/T(\Lambda_1) $$ is an isomorphism. It's surjective because the original map $T$ was surjective to $\mathbb{C}$ so everything in $\Lambda_2$ has a pre-image, which by definition lives in $T^{-1}(\Lambda_2)$. It's injective because the original map $T$ was injective, so if $T(x) \in T(\Lambda_1)$ then $x \in \Lambda_1$. (Having $T(x) \in T(\Lambda_1)$ means there exists $y$ such that $y \in \Lambda_1$ and $T(y) = T(x)$, but now we can conclude $x=y$ by injectivity of the original map $T$.)