Path on a manifold has interval contained inside open ball

Question

Let $M$ be an $m$-dimensional manifold, and let $g:[a,b]\rightarrow M$ be a path. Let $t\in(a,b)$. There exists a neighborhood of $g(t)$ in $M$ that is homeomorphic to an open ball $B(x,\epsilon)$ in $\mathbb{R}^m$, call this map $\phi$. What I wish to show is there there exists a $\delta$ such that $g((t-\delta,t+\delta))\subset B(x,\epsilon)$. Intuitively this makes sense, as I should be able to get a path whose image under my homeomorphism is contained inside this ball, but I can't seem to figure out how to prove that it's true. I need it as part of a larger problem. Does anyone have any hints?

Answer 1

Let $N$ be the neighbourhood of $g(t)$ in $M$. As $N$ is open and $g$ is continuous, $g^{-1}(N)$ is an open subset of $[a, b]$. As $t \in g^{-1}(N)$ and $g^{-1}(N)$ is open, there is $\delta > 0$ such that $(t - \delta, t + \delta) \subseteq g^{-1}(N)$. As such, $g((t - \delta, t + \delta)) \subseteq g(g^{-1}(N)) \subseteq N$.