HI: I have the following question:
Definition: A Lie group $T$ is called a torus if $T\cong \prod_{1\leq i\leq k} \mathbb{R}/\mathbb{Z}$ for some $k\in \mathbb{N}$.
${\bf Question}$: Is it true that a torus is an injective object in the category of abelian Lie groups? Thanks very much!
No; even in the category of connected abelian Lie groups, tori are not injective objects. For ease of notation, we denote by $\mathbb{T}$ the torus $\mathbb{R}/\mathbb{Z}$. Here is a counterexample:
Fix rationally independent, irrational numbers $\xi, \eta$. Let $f : \mathbb{R} \to \mathbb{T}^2$ be given by $$ f : x \mapsto (x\xi, x\eta). $$ Obviously $f$ is an injective map of Lie groups $\mathbb{R} \to \mathbb{T}^2$. Let $\pi : \mathbb{R} \to \mathbb{R}/\mathbb{Z}$ be the quotient map. Then we claim there is no map of Lie groups $h : \mathbb{T}^2 \to \mathbb{R}/\mathbb{Z}$ with $hf = \pi$; this shows that $\mathbb{R}/\mathbb{Z}$ is not an injective object. (It will also follow from this proof that no higher-dimensional torus is either.)
It's well-known (Kronecker's theorem) that every orbit of the map $(x, y) \mapsto (x + \xi, y + \eta)$ is dense in $\mathbb{T}^2$. In particular, there is a sequence $n_k \in \mathbb{Z}$ such that $$ n_k(\xi, \eta) \to \frac{1}{2}(\xi, \eta), $$ from which it follows that $$ \frac{2n_k + 1}{2}(\xi, \eta) \to 0. $$ But then if $h$ is any continuous map $\mathbb{T}^2 \to \mathbb{R}/\mathbb{Z}$ taking $0$ to $0$, $$ (h\circ f)\Big(\frac{2n_k + 1}{2}\Big) = h\Big(\frac{2n_k + 1}{2}(\xi, \eta) \Big) \to 0, $$ whereas $$ \pi \Big(\frac{2n_k + 1}{2}\Big) = \frac{1}{2}, $$ so we cannot have $hf = \pi$.