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Let $M, N$ be Riemannian manifolds. A smooth submersion $f:M \to N$ is called a quasi Riemannian submersion if for every $x\in M$ the restriction of linear map $Df_x$ to orthogonal complement of $\ker Df_x$ pulls back the metric $T_{f(x)} N$ to a scalar multiple of metric of $M$.

1) If a pair $(M,N)$ admit a quasi Riemannian submersion, do they admit a Riemannian submersion too?

2)Let $M$ be a Riemannian manifold and $N$ be a smooth retract of $M$. Do the pair $(M,N)$ admit a quasi Riemannian submersion?

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    $\begingroup$ 1) If I understand the question correctly, then not. Consider $M$ a compact Riemann surface without automoprhisms, let $g$ be the curvature $-1$ metric. Let $g'=\phi g$, where $\phi\ne 1$. Then the identity map: $(M,g)\to (M, \pi g)$ is a quasi-Riemannian submersion. But there is no Riemannian submersion $(M,g)\to (M, \pi g)$, since this would be an isomery, and any isometry gives a non-trivial map $M\to M$ of Riemann surfaces. $\endgroup$
    – Dmitri Panov
    Commented Mar 5, 2020 at 8:56
  • $\begingroup$ @DmitriPanov is not possible that an isometry of $(M,g) \to (M,g')$ fails to be holomorphic, for example orientation reversing? $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2020 at 21:19
  • $\begingroup$ You are right! However, on a generic surface of genus $\ge 1$ there are no orientation reversing conformal self-maps. $\endgroup$
    – Dmitri Panov
    Commented Mar 6, 2020 at 21:58

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