Which closed $n$-manifolds are continuous images of spheres? A closed $n$-manifold is taken to mean a compact topological $n$-manifold without boundary.
I have been thinking about (higher) homotopy groups and possible analogues, and while I have good intuition for closed curves being continuous images of the circle*, my intuition rapidly decreases with increasing dimension. Still, I would like to keep the question as intuitive as possible, which is why I narrowed it down to (topological) manifolds.

Question 1: Are there closed $n$-manifolds $M^n$, $ n \geq 3$, that are not continuous images of $\mathbb{S}^n$? If so, can we nevertheless find $m > n$ such that $M^n$ is a continuous image of $\mathbb{S}^m$? (In other words, are there closed $n$-manifolds that are not continuous images of any spheres?)

For example, IIRC there are many surjective maps $\mathbb{S}^2 \to \Sigma_g$, where $\Sigma_g$ is an orientable closed surface of genus $g$ (needs a reference). Also, for non-orientable closed surfaces, I guess we can start with the covering $S^2 \to \mathbb{RP}^2$.

Question 2: Is there a classification (up to homotopy equivalence) of closed $n$-manifolds, $n \geq 3$, that are continuous images of $\mathbb{S}^n$? If not, is there a reason why such a classification could be deemed hopeless? Can we improve on that by assuming smoothness throughout?

*Well, strictly speaking, closed curves are continuous maps from the circle and their images are called traces of the closed curves, but I hope you will forgive me the abuse of terminology.
 A: Each connected compact $n$-manifold (with our without boundary) is the continuous image of $S^m$ for all $m \ge 1$. To see this, note that $\phi : S^m \to I = [0,1], \phi(x_1,\ldots,x_{m+1}) = \lvert x_1 \rvert$, is a continuous surjection.
$M$ can be covered by finitely many closed $B_1,\ldots, B_k \subset M$ which are homeomophic to the cube $I^n$. It is well-known that there exists a continuous surjection $f_2 : I \to I^2$. By iteration we get continuous surjections $f_n : I \to I^n$ and thus continuous surjections $f^i_n : I \to B_i$. Hence there exist continuous surjections $g_i : J_i = [\frac{2i-2}{2k},\frac{2i-1}{2k} ] \to B_i$. Choose paths $u_i : L_i = [\frac{2i-1}{2k},\frac{2i}{2k} ] \to M$ such that $u_i(\frac{2i-1}{2k}) = b_i = g_i(\frac{2i-1}{2k})$ and $u_i(\frac{2i}{2k}) = a_{i+1} = g_{i+1}(\frac{2i}{2k})$, $i = 1,\ldots, k$, where we set formally $a_{k+1} = b_k$.
Define
$$\psi : I \to M, \psi(t)  = \begin{cases} g_i(t) & t \in J_i \\u_i(t) & t \in L_i \\  \end{cases} $$
This is a continuous surjection, hence also $\psi \circ \phi$ is one.
A: You can also argue like so I believe: endow $M$ with a riemannian metric $g$ and pick a point $m\in M$. Unless I'm mistaken the restriction of the exponential map $\exp_{m}$ to the closed ball $B$ centered at $0$ of radius $\mathrm{diam}_g(M)$ is a smooth surjection $B\to M$. You thus get an onto map $S^n\to T_mM\to M$ where the first map is the projection $S^n\to B\subset T_mM$ parallel to the $(n+1)$-th coordinate (+ scaling by $\mathrm{diam}_g(M)$).
