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But paragraph which I marked by red line seems to me confusing.

Let $k=3$ then $\sigma=[\mathbf{p}_0,\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3]$ and $\partial \sigma=[\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3]-[\mathbf{p}_0,\mathbf{p}_2,\mathbf{p}_3]+[\mathbf{p}_0,\mathbf{p}_1,\mathbf{p}_3]-[\mathbf{p}_0,\mathbf{p}_1,\mathbf{p}_2].$ Since $T:E\to V$ is affine then $T(\mathbf{x})=T(a)+A\mathbf{x}$ for some $a\in E$ and $A\in L(\mathbb{R}^n,\mathbb{R}^m)$.

1) We will show that $T\circ \sigma$ is oriented affine $k$-simplex. $$T\circ \sigma(u)=T(a)+A(\mathbf{p}_0+u_1(\mathbf{p}_1-\mathbf{p}_0)+u_2(\mathbf{p}_2-\mathbf{p}_0)+u_3(\mathbf{p}_3-\mathbf{p}_0))=$$$$=[T(a)+A\mathbf{p}_0,T(a)+A\mathbf{p}_1,T(a)+A\mathbf{p}_2,T(a)+A\mathbf{p}_3];$$

2) We have to show that $\partial (T\circ \sigma)=T(\partial \sigma)$.

$$\partial (T\circ \sigma)=[T(a)+A\mathbf{p}_1,T(a)+A\mathbf{p}_2,T(a)+A\mathbf{p}_3]-[T(a)+A\mathbf{p}_0,T(a)+A\mathbf{p}_2,T(a)+A\mathbf{p}_3]+[T(a)+A\mathbf{p}_0,T(a)+A\mathbf{p}_1,T(a)+A\mathbf{p}_3]-[T(a)+A\mathbf{p}_0,T(a)+A\mathbf{p}_1,T(a)+A\mathbf{p}_2]$$ by definition 10.29.

But $$T(\partial \sigma)=T(a)+A(\mathbf{p_1}+u_1(\mathbf{p_2-\mathbf{p_1}})+u_2(\mathbf{p_3}-\mathbf{p_1}))-A(\mathbf{p_0}+u_1(\mathbf{p_2-\mathbf{p_0}})+u_2(\mathbf{p_3}-\mathbf{p_0}))+A(\mathbf{p_0}+u_1(\mathbf{p_1-\mathbf{p_0}})+u_2(\mathbf{p_3}-\mathbf{p_0}))-A(\mathbf{p_0}+u_1(\mathbf{p_1-\mathbf{p_0}})+u_2(\mathbf{p_2}-\mathbf{p_0}))=T(a)+\partial(T\circ \sigma)$$ and we see that $T(\partial \sigma)\neq\partial(T\circ \sigma)$.

Can anyone please explain what's wrong with my reasoning?

RFZ
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    In 1), you're mixing things. In the first line, you look at $(T\circ \sigma)(u)$, but in the second line you have the representation of an affine simplex by the (ordered) list of its vertices. Note also that $T(a) + A\mathbf{p}_i = T(\mathbf{p}_i)$, so for $\sigma = [\mathbf{p_0}, \mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_3]$ you get $T\circ\sigma = [T(\mathbf{p}_0), T(\mathbf{p}_1), T(\mathbf{p}_2), T(\mathbf{p}_3)]$. Now, for $T(\partial \sigma)$, you need to use the definition of $T\circ \Gamma$ for affine chains (because $\partial \sigma$ is an affine chain). And that gives – Daniel Fischer May 03 '16 at 22:00
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    $$\begin{aligned}T(\partial \sigma) &= T\bigl([\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3] - [\mathbf{p}_0, \mathbf{p}_2, \mathbf{p}_3] + [\mathbf{p}_0, \mathbf{p}_1, \mathbf{p}_3] - [\mathbf{p}_0, \mathbf{p}_1, \mathbf{p}_2])\ &=T\circ [\mathbf{p}_1, \mathbf{p}_2, \mathbf{p}_3] - T\circ [\mathbf{p}_0, \mathbf{p}_2, \mathbf{p}_3] + T\circ [\mathbf{p}_0, \mathbf{p}_1, \mathbf{p}_3] - T\circ [\mathbf{p}_0, \mathbf{p}_1, \mathbf{p}_2]\end{aligned}$$ – Daniel Fischer May 03 '16 at 22:02
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    $$\begin{aligned}\hphantom{T(\partial \sigma)}&= [T(\mathbf{p}_1), T(\mathbf{p}_2), T(\mathbf{p}_3)] - [T(\mathbf{p}_0), T(\mathbf{p}_2), T(\mathbf{p}_3)] + [T(\mathbf{p}_0), T(\mathbf{p}_1), T(\mathbf{p}_3)] - [T(\mathbf{p}_0), T(\mathbf{p}_1), T(\mathbf{p}_2)]\ &= \partial [T(\mathbf{p}_0), T(\mathbf{p}_1), T(\mathbf{p}_2), T(\mathbf{p}_3)]\ &= \partial (T\circ \sigma). \end{aligned}$$ – Daniel Fischer May 03 '16 at 22:02
  • @DanielFischer, I guees that you made a typo: $T(\mathbf{p}_i)=T(a)+A\mathbf{p}_i$. Right? Also you mentioned that I made mistake in my first and second lines but unfortunately i didn't find no mistake. Can you explain this in detail please? – RFZ May 04 '16 at 08:05
  • Right, that was a typo, thanks. Corrected now. (Being a moderator has some advantages ;-) In 1), your chain of equations is structurally $(f\circ g)(u) = h(u) = h$, where $f,g,h$ are functions. $h(u)$ and $h$ are things of different kinds, so $h(u) = h$ is incorrect. That's what I meant with "mixing things". If you make the last expression $T(a) + A\mathbf{p}_0, \dotsc$ it will be correct. – Daniel Fischer May 04 '16 at 09:33
  • Thanks a lot for your always brilliant answer! If you have a free time may I ask you question about this? http://math.stackexchange.com/questions/1771170/positively-oriented-boundaries-from-pma-rudin – RFZ May 04 '16 at 11:24
  • @DanielFischer, Unfortunately Rudin misses any hint or comment how to conclude it :( – RFZ May 04 '16 at 11:43

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