Let $\gamma(t)=(r\cos t, r\sin t)$ where $r>0$ is fixed and $t\in [0,2\pi]$. Rudin write that it's an "oriented 1-simplex". Also he states that $$\partial \gamma=0.$$ Let $T(u)=(r\cos u,r\sin u)$ and $\sigma=[0,2\pi]$. We see that $\sigma(u)=2\pi u$ for $u\in Q^1=[0,1]$. We see that $T\in C''$ and $T\circ \sigma(u)$ for $u\in Q^1$ is circle with radius $r>0$ on the plane. Hence by definition 10.30 $\Phi=T\circ \sigma$ is oriented 1-simplex. Right? By definition of boundary $$\partial \gamma=\partial (T\circ \sigma)=T(\partial \sigma)=T([2\pi]-[0])=?$$
How to show that above expression is zero?
P.S. It's from Rudin PMA example 10.36.
Can anyone please help with this moment?
By definition of $T(\Gamma)$ where $\Gamma$ is a $k$-chain (here $k = 0$), we have
$$T([2\pi] - [0]) = T([2\pi]) - T([0]).$$
For the given $T$, we have
$$T([2\pi]) = [(r,0)] = T([0]),$$
and thus
$$\partial \gamma = [(r,0)] - [(r,0)] = 0.$$
Generally, if $\alpha \colon [a,b] \to \mathbb{R}^n$ is a (sufficiently smooth) curve, we have $\partial \alpha = [\alpha(b)] - [\alpha(a)]$. If the curve is closed, that reduces to $0$.