Conjugation is continuous

Idea. Write the map as a composite of maps that are already known to be continuous.

Step 1. The map $(P,A)\mapsto(P,A,P^{-1})$ from $GL_n\times M_n$ to $M_n\times M_n\times M_n$ is continuous: its components are the two projections and the inversion $P\mapsto P^{-1}$ (Problem 68). (A map into a product is continuous iff its components are.)

Step 2. Multiplication $M_n\times M_n\to M_n$ is continuous (Problem 67), so $(P,A,P^{-1})\mapsto PA$ and then $(PA,P^{-1})\mapsto(PA)P^{-1}$ are continuous.

Step 3. Composing, $(P,A)\mapsto PAP^{-1}$ is continuous.

Consequences.
- For fixed $P$, $A\mapsto PAP^{-1}$ is a linear automorphism of $M_n$, hence a homeomorphism that preserves trace, determinant, rank, characteristic polynomial and spectrum.
- The *similarity class* (orbit) $\{PAP^{-1}\}$ is the image of $GL_n$ under the continuous map $P\mapsto PAP^{-1}$.

Remark. The orbits are generally not closed: the orbit of $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ contains $\begin{pmatrix}1&t\\0&1\end{pmatrix}$ for all $t\ne0$ (conjugate by $\operatorname{diag}(t,1)$), and $t\to0$ gives $I$, which is not in the orbit. Compare the analysis of closures of similarity classes in Part 6.