# If $f: M \to N$ is a diffeomorphism and $N$ is complete, then $M$ is complete

This is a problem from Riemannian Geometry by Do Carmo, namely Ch. 7, Sec. 3, Problem 7 on pg. 153.

Let $M, N$ be Riemannian manifolds with $N$ complete, and $f: M \to N$ a diffeomorphism for which there exists a $c > 0$ such that $|v| \geq c |df_p(v)|$ $\forall p \in M$ $\forall v \in T_p M$.

I figure I need to apply the Hopf-Rinow theorem or just go from the definition and show $exp_p$ is defined on all of $T_p M$, i.e. that any geodesic $\gamma(t)$ starting from $p$ is defined $\forall t \in \mathbb{R}$. I don't really know how to go about this, though.

It is easier to apply the following Hopf Rinow theorem:

$(N, h)$ is geodesically complete if and only if the following metric $$d_N (r, s) := \inf\left\{ \int_0^1 |\dot \gamma(t)|_h dt \bigg| \gamma :[0,1] \to N, \gamma(0)= r, \gamma(1) = s\right\}$$ is a complete metric on $N$.

Together with your condition, one has (check!)

$$\tag{1} d_N (f(p), f(q)) \le \frac 1c d_M(p, q),\ \ \ \forall p,q\in M.$$

Now we can claim:

$(M, d_M)$ is complete.

Let $\{p_n\}$ be a Cauchy sequence in $M$. By $(1)$, $\{f(p_n)\}$ is Cauchy in $N$ and so $f(p_n) \to Q \in N$ by completeness of $N$. Note that there is $C >0$ and a closed ball $B_Q$ around $Q$ (since $f$ is a diffeomorphism) so that

$$|v| \le C | df_q v|,\ \ \ \forall f(q) \in B_Q.$$

Thus for large $n$ we have $f(p_n) \in B_Q$ and so

$$d_M (p_n, f^{-1}(Q))\le Cd_N(f(p_n), Q)$$

Thus $p_n \to f^{-1}Q$ and so $(M, d_M)$ is complete.

• I think the constant in your formula (1) is 1/c. – Yilong Zhang Mar 21 '16 at 23:06
• @YilongZhang : Yes, thanks and edited. – user99914 Mar 21 '16 at 23:08
• Where did you get this Hopf-Rinow theorem? Do Carmo has a different version, and I wanted to restrict myself to his machinery. Is there a way to do it differently? Hmm, upon further inspection, this would seem to follow from his Proposition 2.6, right? – Ryker Mar 22 '16 at 1:31
• That's Theorem 2.8 (c) (second printing 1993 version) in the book. @Ryker – user99914 Mar 22 '16 at 1:35
• Ah, so the statement that $M$ is complete as a metric space? – Ryker Mar 22 '16 at 1:38