Let $G$ act smoothly on $M$, $(g,m) \rightarrow t_g (m) $ from the left. Why is $t_g \simeq id $ if $G$ connected?

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Let $G$ act smoothly on $M$, $(g,m) \rightarrow t_g (m) $ from the left. Why is $t_g \simeq id $ if $G$ connected? I've seen this statement, but have no idea how to use connectedness here. Does anyone has hints?

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$G$ is connected implies it is connected by arcs, you have a path $c:[0,1]\rightarrow G$ such that $c(0)=e, c(1)=g$, you can define $H(t,x)=t_{c(t)}(x)$. $H(0,x)=t_{c(0)}x=x$ and $H(1,x)=t_{c(1)}x=t_g(x)$.