Unexpected isomorphism

Since Cantor's "I see it, but I cannot believe it" (1877), we know that \(\mathbb{R}^n\) are isomorphic sets for all \(n > 0\). This being as shocking as it is, over time we learn to live with it, because the bijections between continua of different dimensions are extremely discontinuous and we assume that if we limit ourselves to any reasonably well-behaving class of maps the isomorphisms will disappear. Will they?

Theorem. Additive groups \(\mathbb{R}^n\) are isomorphic for all \(n > 0\) (and, therefore, isomorphic to the additive group of the complex numbers).

Proof. Each \(\mathbb{R}^n\) is a vector space over rationals. Assuming axiom of choice, any vector space has a basis. By simple cardinality considerations, the cardinality of a basis of \(\mathbb{R}^n\) over \(\mathbb{Q}\) is the same as cardinality of \(\mathbb{R}^n\). Therefore all \(\mathbb{R}^n\) have the same dimension over \(\mathbb{Q}\), and, therefore, are isomorphic as vector spaces and as additive groups. End of proof.

This means that for any \(n, m > 0\) there are bijections \(f : \mathbb{R}^n \to \mathbb{R}^m\) such that \(f(a + b) = f(a) + f(b)\) and, necessary, \(f(p\cdot a + q\cdot b) = p\cdot f(a) + q\cdot f(b)\) for all rational \(p\) and \(q\).

I feel that this should be highly counter-intuitive for anybody who internalised the Cantor result, or, maybe, especially to such people. The reason is that intuitively there are many more continuous maps than algebraic homomorphisms between the "same" pair of objects. Indeed, the formula defining continuity has the form \(\forall x\forall\epsilon\exists\delta\forall y P(x, \epsilon, \delta, y)\) (a local property), while homomorphisms are defined by \(\forall x\forall y Q(x, y)\) (a stronger global property). Because of this, topological categories have much denser lattices of sub- and quotient-objects than algebraic ones. From this one would expect that as there are no isomorphisms (continuous bijections) between continua of different dimensions, there definitely should be no homomorphisms between them. Yet there they are.