Suppose $M$ is an $m$-dimensional closed Riemannian manifold and $f \colon M \to \mathbb{R}^{n}$ is continuous. I'm interested in the case when $M=\mathbb{T}^{m}$. Let $\mathcal{M}_{f} := f(M)$. Let $d$ be the box dimension of $\mathcal{M}_{f}$, i. e. $$d=\lim_{\varepsilon \to 0}\frac{\ln N(\varepsilon)}{\ln 1/\varepsilon}=\inf\{ s\geq 0 \ | \ \lim_{\varepsilon \to 0} \varepsilon^{s} N(\varepsilon) = 0 \},$$ where $N(\varepsilon)$ is the least number of balls of radius $\varepsilon$ required to cover $\mathcal{M}_{f}$. Put $V_{B}:=\lim_{\varepsilon \to 0} \varepsilon^{d}N(\varepsilon)$.

For the following questions one may assume that $M=\mathbb{T}^{m}$ and even $m=2$, $n \geq m$.

**Question 1.** Is it true that the values $d$ and $V$ are well-defined for $\mathcal{M}_{f}$?

**Question 2.** Is it true that $d=\dim_{H}\mathcal{M}_{f}$, where $\dim_{H}$ is the Hausdorff dimension.

The answer to **Q.2** is positive in the case $f$ is smooth and has at least one regular point, since in such a case we have $m \leq \dim_{T}\mathcal{M}_{f}\leq \dim_{H}\mathcal{M}_{f} \leq \overline{\dim}_{B}\mathcal{M}_{f} \leq m$.

I will be grateful for any help.