Let $f:A\to A'$ be a morphism of elliptic curves over the real numbers $\mathbb R$. It canonically induces a morphism $f(\mathbb R): A(\mathbb R)\to A'(\mathbb R)$ between the sets of real points, which carry a structure of a real Lie group.
Is $f(\mathbb R)$ a morphism of real Lie groups?
Does this hold for arbitrary abelian varieties $A,A'$?
Let $A$, $A'$ be abelian varieties over $\mathbb{R}$, and let $f:A\to A'$ be a morphism. If $f(0_A)\neq 0_{A'}$, then we define $\widehat{f}:A\to A'$ as $\tau_P\circ f'$ where $\tau_P(Q)=Q-P$, and $P=f(0_A)$. Then, $\widehat{f}$ is a morphism and $\widehat{f}(0_A)=0_{A'}$.
In addition, we require here that $f$ is surjective with finite fibres (alternatively, require that $f$ is a dense morphism - see the discussion here), so that $\widehat{f}$ is also a group homomorphism.
Finally, let $Q\in A(\mathbb{R})$ be fixed. Since $\widehat{f}$ is a morphism, the function $\widehat{f}$ is regular at $Q$. In particular, locally at $Q$ the function $\widehat{f}$ is defined by rational functions whose denominators do not vanish at $Q$. Since rational functions are meromorphic (with poles only at zeros of the denominators, if any), it follows that $\widehat{f}$ is in fact smooth at $Q$. Thus, $\widehat{f}$ is a smooth group homomorphism of real Lie groups, and therefore a morphism of real Lie groups.