Metric induced by submersion If $(N,h)$ is a Riemannian manifold and $F:M\to N$ is an immersion, we can trivially make $F$ into an isometric immersion by placing the induced metric on $M$. My question concerns the analogous procedure for submersions. 
Let $f:M\to N$ be a submersion and let $g$ be a Riemannian metric on $M$. Let $\mathcal{V,H}$ be the vertical and horizontal distributions, respectively, i.e. the tangent space of $M$ at $p$ decomposes as $T_p M=\mathcal{V}_p\oplus \mathcal{H}_p$ for all $p\in M$. The restriction $\varphi_p:=(d\pi_p)|_{\mathcal{H}_p}:\mathcal{H}_p\to T_{f(p)}N$ of the differential is an isomorphism for all $p\in M$. 
Is there a way to construct a metric on $N$ so that $f$ becomes a Riemannian submersion?
My idea is to define a metric $h$ on $N$ by $$h_{f(p)}(X,Y)=g_p(\varphi_p^{-1}(X),\varphi_p^{-1}(Y))?$$ Does this work?
 A: I think that in general, there is no metric on $N$ for which $\pi$ becomes a Riemannian subermsion.  The issue is that you want $d\pi_p:\mathcal{H}_p\rightarrow T_{f(p)}$ to be an isometry for every $p$.  In particular, if $p,q$ have $f(p)=f(q)$, then you need $(d\pi_q)^{-1} (d\pi_p):\mathcal{H}_p\rightarrow \mathcal{H}_q$ to be an isometry, and there is no reason it should be.
For an explicit example, let $M$ be the the cylinder $M = S^1\times \mathbb{R}$ with the natural projection map $\pi:M\rightarrow N:=S^1$.  Suppose the metric on $M$ is a warped product metric of the form $\phi(t)d\theta^2 + dt^2$.  Then the horizontal space at $p\in M$ is spanned by $\frac{\partial}{\partial \theta}$.
Then $M\rightarrow N$ is a Riemannian submersion for some metric on $N$ iff $\phi$ is constant.
(On the other hand, if the Riemannian metric on $N$ is assigned, I believe there is a metric on $M$ which makes $M\rightarrow N$ a Riemannian submersion.  To do this, pick any random metric on $M$ in order to talk about $\mathcal{V}_p$ and $\mathcal{H}_p$.  Then change this random metric on $\mathcal{H}_p$ by declaring $d\pi_p$ to be an isometry).
