Riemannian manifolds and lipschitz maps Consider a connected Riemannian manifold $(M,d)$ where $d$ is the induced metric and $Y$ is some metric space.
Suppose $f:(M,d)\rightarrow Y$ is locally $K$ Lipschitz. Why does it follow that $f$ is globally Lipschitz?
I was told that this can be answered through a purely metric sense.
My attempt so far:
I suppose if $L(\gamma)$ for every curve $\gamma$ from every $x$ to $x'$   is bounded above by $Cd(x,x')$  then by a Lemma in class called $C-quasiconvexity$ Lemma, the result follows. Though I am not sure how to use that.
 A: You're not going to be able to prove that every curve $L(\gamma)$ from $x$ to $x'$ has length bounded above by $C \, d(x,x')$, given that you can replace any small segment of any curve from $x$ to $x'$ by a nonrectifiable curve (i.e. one of infinite length).
Instead, by definition of the metric $d(x,x')$, for any $\epsilon > 0$ you may connect $x$ to $x'$ by a path $\gamma : [0,1] \to M$ of length $L(\gamma) \le d(x,x') + \epsilon$.
Knowing that $f$ is locally $K$-Lipschitz, choose a covering $\{U_i\}$ of open subsets of $M$ such that $f \mid U_i$ is $K$-Lipschitz.
Consider the open covering $\{\gamma^{-1}(U_i)\}$ of $[0,1]$. By the Lebesgue number lemma we can choose $\lambda>0$ such that if $s,t \in [0,1]$ and $|s-t|<\lambda$ then $s,t \in \gamma^{-1}(U_i)$ for some $i$, hence $\gamma(s),\gamma(t) \in U_i$, hence $f(\gamma(s),\gamma(t)) \le K \, d(\gamma(s),\gamma(t))$. Choose a subdivision
$$0 = s_0 < s_1 < ... < s_n = 1
$$
into subintervals $[s_{i-1},s_i]$ of length $<\lambda$, so
\begin{align*}
d(f(x),f(x')) &\le d(f(\gamma(0)),f(\gamma(1))) \\
        &\le d(f(\gamma(s_0)),f(\gamma(s_1))) + \cdots + d(f(\gamma(s_{n-1})),f(\gamma(s_n))) \\
        &\le K \cdot \bigl(d(\gamma(s_0),\gamma(s_1)) + \cdots + d(\gamma(s_{n-1}),\gamma(s_n))\bigr) \\ 
        &\le K \cdot L(\gamma) \\
        &\le K \cdot (d(x,x') + \epsilon)
\end{align*}
Letting $\epsilon \to 0$ you get
$$d(f(x),f(x')) \le K \, d(x,x')
$$
