Diffeomorphism and homeomorphism between open ball and polydisc in $\mathbb{R}^n$

I know that the product topology on $\mathbb{R}^n$ is equivalent to the metric topology, i.e. topology generated by open polydiscs is same as the one generated by open balls. Does this imply that $B(0,1)$ is homeomorphic to $\Delta(0;1)$?

Not if it's for a topological reason. For example, the topology on $\mathbb R^n$ is also generated by the sets which are unions of two open balls of radius $r$ and whose centers are at distance $5r$. But a disjoint union $B(0, 1) \sqcup B(0, 1)$ is definitely not homeomorphic to $B(0, 1)$.

• We have a smooth map $$\begin{align*} B(0, 1) & \to \mathbb R^n \\ x & \mapsto \frac{x}{\sqrt{1- \|x\|^2}} \end{align*}$$ with smooth inverse $$\begin{align*} \mathbb R^n & \to B(0, 1) \\ y & \mapsto \frac{y}{\sqrt{1 + \|y\|^2}} \end{align*}$$

• If $g : (-1, 1) \to \mathbb R$ is any diffeomorphism, then $$\begin{align*} \Delta(0, 1) & \to \mathbb R^n \\ (x_1, \ldots, x_n) & \mapsto (g(x_1), \ldots, g(x_n)) \end{align*}$$ is a smooth map with smooth inverse $$\begin{align*} \mathbb R^n & \to \Delta(0, 1) \\ (y_1, \ldots, y_n) & \mapsto (g^{-1}(y_1), \ldots, g^{-1}(y_n)) \end{align*}$$

They are diffeomorphic since they are both open star shaped sets, see A star shaped open set in $\mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^n$

You can indeed show that two balls for two (equivalent) norms are homeomorphic :

Homeomorphism between open unit balls from equivalent distances