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I was recently interested in the following conjecture, which at first sight seemed pretty elementary.

Conjecture: Let $i: A \hookrightarrow B$ be an injection into a finitely generated abelian group. Then the induced map $i^*: Hom(B,U(1)) \rightarrow Hom(A,U(1))$ is surjective.

(Note that this conjecture is not very natural, in that the universal property of monomorphism only tells us the induced map $Hom(C,A)\rightarrow Hom(C,B)$ is injective.)

So far I've only proved a weaker version where $B$ is finitely generated free abelian, as laid out below. However I'm not sure how to ensure the $g_i$'s in the final proposition to have the same orders as the generators $a_i$ of $B$ in case $B$ is not free. Maybe there's a proof of the conjecture somewhere in the literature that I wasn't aware of? Or is it actually false and I was being stupid?

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Lemma 1. If $\Lambda$ is a discrete subgroup of $\mathbb R^n$, then $\Lambda$ is a lattice in the sub-vector space $V\subset \mathbb R^n$ generated by $\Lambda$. That is, $\Lambda=\mathbb Z v_1 + \ldots + \mathbb Z v_d$ for a basis $\{v_1, \ldots, v_d\}$ of $V$.

Lemma 2. Let $M$ be an integer-valued $k\times l$ matrix. Suppose the rows of $M$ are linearly dependent (over $\mathbb R$). Then, there exist some $r<k$, some integer-valued $r\times l$ matrix $N$ with linearly independent rows, and some integer-valued $k\times r$ matrix $\Gamma$, such that $M = \Gamma N$.

Lemma 3. Let $M=(m_{ij})$ be an integer-valued $k\times l$ matrix with linearly independent rows. Let $f_1, \ldots, f_k \in U(1)$. Then there exist $g_1, \ldots, g_l\in U(1)$ such that $g_1^{m_{i1}} g_2^{m_{i2}} \ldots g_l^{m_{il}} = f_i$ for all $i=1, \ldots, k$.

Proof: This would be elementary linear algebra if $U(1)$ were replaced by $\mathbb R$. To prove the $U(1)$ case one just needs a little extra care. W.l.o.g. we can assume $M = [M_1 ~ M_2]$ where $M_1$ is an invertible square matrix. Choose $h_i$ s.t. $h_i^{\det M_1} = f_i$ for each $i$. Let $N = (n_{ij})$ be the integer-valued square matrix $(\det M_1) M_1^{-1}$. Let \begin{equation} g_i = \begin{cases} h_1^{n_{i1}} \ldots h_k^{n_{ik}}, & 1\leq i \leq k, \\ 1, & i >k. \end{cases} \end{equation} The desired equation is easily verified to be satisfied: \begin{eqnarray} g_1^{m_{i1}} \ldots g_l^{m_{il}} &=& (h_1^{n_{11}} \ldots h_k^{n_{1k}})^{m_{i1}} \ldots (h_1^{n_{k1}} \ldots h_k^{n_{kk}})^{m_{ik}} \\ &=& h_1^{n_{11} m_{i1} + \ldots + n_{k1} m_{ik}} \ldots h_k^{n_{1k} m_{i1} + \ldots + n_{kk} m_{ik}} \\ &=& h_1^{\delta_{i1} \det M_1} \ldots h_k^{\delta_{ik} \det M_1} \\ &=& f_i. \end{eqnarray}

Proposition 4. Let $i:A \hookrightarrow B$ be an injection into a finitely generated free abelian group. Let $f:A \rightarrow U(1)$ be a homomorphism. Then there exists a homomorphism $g:B \rightarrow U(1)$ such that $g|A = f$.

Proof: Let $a_1, \ldots a_l$ be the generators of the $\mathbb Z$-factors of $B \cong \mathbb Z \oplus \ldots \oplus \mathbb Z$. Let $x_1, \ldots, x_k$ be a set of generators of $A$. Then, \begin{equation} \begin{bmatrix}x_1 \\ \vdots \\x_k\end{bmatrix} = \begin{bmatrix} m_{11} & \ldots & m_{1l} \\ \vdots & & \vdots \\ m_{k1} & \ldots & m_{kl} \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_l \end{bmatrix} \end{equation} for some integer-valued matrix $M =(m_{ij})$. By Lemma 2 we can assume $M$ has linearly independent rows. Then, by Lemma 3 there exist $g_1, \ldots, g_l\in U(1)$ s.t. $g_1^{m_{i1}} g_2^{m_{i2}} \ldots g_l^{m_{il}} = f(x_i)$ for all $i=1, \ldots, k$. Now we can define $g:B\rightarrow U(1)$ by setting $g(a_i) = g_i$. This completes the proof.

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  • $\begingroup$ $U(1)$ refers to the first unitary group, i.e. $(S^1,*,1)$? $\endgroup$ – Martin Brandenburg Apr 8 '15 at 21:31
  • $\begingroup$ Yes, that's right. $\endgroup$ – user46652 Apr 8 '15 at 22:03
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This is true without any f.g. assumption and is equivalent to the statement that the circle group $U(1)$ is injective, which follows from Baer's criterion since it is divisible. If your groups are f.g., then you won't need Zorn's Lemma in Baer's criterion (it can be replaced by a simple induction).

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  • $\begingroup$ Got you. So $Hom(-,U(1))$ is an exact functor, since $U(1)$ is injective. Thanks! $\endgroup$ – user46652 Apr 8 '15 at 22:32

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