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Suppose we have a Riemannian manifold $(M^m, g_M)$ and a smooth manifold $N^n$ (with dimensions $m > n$). And suppose we have a smooth surjective submersion $F: M^m \to N^n$.

Question: Is there a natural way to construct a Riemannian metric $g_N$ on $N^n$ such that $F$ can actually be "upgraded" to a Riemannian submersion?

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In most cases, there is no Riemannian metric on $N$ which makes $F$ a Riemannian submersion. However, if there is such a metric, it is unique. To shed more light on this matter, let us try to construct the desired metric.

Let $p\in N$, and let $q\in F^{-1}(p)$. Let $H_q$ denote the orthogonal complement of $\ker dF_q$ in $T_qM$. Then $dF_q$ restricts to an isomorphism of vector spaces $$dF_q|_{H_q}:H_q\xrightarrow{\sim}T_pN,$$ and there is a unique inner product on $T_pN$ which makes the above isomorphism an isometry. This explains the uniqueness part in the first paragraph. Now, if we repeat this argument for some $q'\in F^{-1}(p)$ other than $q$, there is no reason to expect we get the same inner product on $T_pN$, and this is why the desired Riemannian metric does not exist usually.

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    $\begingroup$ (+1) It may be worth mentioning that there's an important but very special case where everything does work out: a $G$-invariant Riemannian metric on the total space $P$ of a principal $G$-bundle $P \to B$ canonically induces a unique Riemannian metric on the base space $B$ that makes the quotient map $P \to B$ a Riemannian submersion. $\endgroup$ – Branimir Ćaćić Dec 22 '17 at 23:37
  • $\begingroup$ Clear explanation. Thanks! $\endgroup$ – user32416 Dec 23 '17 at 2:37
  • $\begingroup$ @AmitaiYuval, thanks for good explanation. I am aware that it is not appropriate to ask another question in comments, but I want to know about your last sentence by a concrete example. $\endgroup$ – C.F.G Aug 15 at 18:49

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