My question is "inspired" by the [uniformization theorem for Riemmannian surfaces](https://en.wikipedia.org/wiki/Uniformization_theorem) and [this post](https://mathoverflow.net/questions/45953/is-every-real-n-manifold-isomorphic-to-a-quotient-of-mathbbrn).




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Suppose that $X$ is connected (finite-dimensional) topological manifold without boundary. Does there exist $n\_{i,j}\in \mathbb{N}\times \{1,2,3\}$, $K\_i\in\mathbb{N}$, and finite isometry groups $\Gamma\_{i,1}$ acting on $S^{n\_k}$, $\Gamma\_{i,2}$ on $\mathbb{R\_{n\_{k,2}}}$ and $\Gamma\_{n\_{k,3}}$ acting on the hyperbolic space $\mathbb{H}^{n\_{k,3}}$
$$
X 
\cong
\prod\_{k=1}^{K\_1} S^{n\_{k,1}}/\Gamma\_{k,1}
\times 
\prod\_{k=1}^{K\_2} \mathbb{R}^{n\_{k,2}}/\Gamma\_{k,2},
\times 
\prod\_{k=1}^{K\_3} \mathbb{H}^{n\_{k,3}}/\Gamma\_{k,3}?
\label{1}\tag{1}
$$
*Here $\cong$ denotes the existence of a homeomorphism.*




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If not, what can we say about all objects on the right-hand side of \eqref{1}?



Can every manifold be represented as a quotient
The answer to your first question is “no”. The first counter-examples appear in dimension four - for example the complex projective plane. There are many simply connected manifolds and that are not aspherical that will fit the bill.


Here is a somewhat wilder example. Suppose that $V$ is a genus two handlebody of dimension three. For example, $V$ can be obtained by embedding a “eye-glasses” graph in three-space and taking a small regular neighbourhood. Let $M$ be the *double* of $V$ across its boundary. That is, we take two copies of $V$ and glue via the identity on the boundary.


The universal cover of $M$ is a copy of the three-sphere, minus a Cantor set.


I don’t see a path to an answer to your second question. Perhaps you would be interested in Thurston’s geometrisation programme. It is sometimes described as a version of uniformisation in dimension three. It sadly does *not* generalise to dimension four.


