Is every automorphism of a topological group continuous with regard to its own topology?

Question

Is every automorphism of a topological group continuous with regard to its own topology?

Note that its inner automorphism obviously is continuous.

If the conclusion hold, then it follows that the connected component of a topological group containing the unit is a characteristic subgroup.

Answer 1

If $K/\mathbb{Q}$ is a number field (finite extension), any automorphism of $K$ can be extended to an automorphism of $\mathbb{C}$. This is the standard way of showing that $\mathbb{C}$ has infinitely many field automorphisms. For example, there's an automorphism of $\mathbb{C}$ which swaps $\sqrt{2}$ and $-\sqrt{2}$.

On the other hand, any continuous automorphism of $\mathbb{C}$ must keep $\mathbb{R}$ pointwise fixed, so it must either be the identity or complex conjugation.

Therefore, the topological groups $(\mathbb{C},+,0)$ and $(\mathbb{C},\cdot,1)$ both have uncountably many discontinuous group automorphisms.

Answer 2

It seems that it depends whom you ask: Pontryagin in his Topological Groups treatise (Pontryagin, L. S., Brown, A., & Naidu, P. S. V. (2018). Topological groups. Routledge) and Bourbaki in his General Topology book (Bourbaki, N. (2013). General Topology: Chapters 1–4 (Vol. 18). Springer Science & Business Media) require any automorphism of a topological group to be a homeomorphism. On the other hand Hewitt and Ross (Hewitt, E., & Ross, K. A. (2012). Abstract Harmonic Analysis: Volume I Structure of Topological Groups Integration Theory Group Representations (Vol. 115). Springer Science & Business Media) distinguish between automorphisms and topological automorphisms of a topological group. The latter being a homeomorphism.