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I´ve a question about singular homology.

Theorem 29.6 in Kosniowski's book uses $\delta_j \delta_i = \delta_i \delta_{j+1}$, if $i \leq j$. My question is for the particular case $i=0 \leq j = 1 \leq n=2$ we have, for $\phi$ a singular $n$-simplex:

$$\delta_1 \delta_0(\phi) = \delta_0 (\phi(x_0,0,x_1)) = \phi(0,x_0,0,x_1)$$

$$\delta_0\delta_2(\phi) = \delta_0 (\phi(x_0,x_1,0)) = \phi(0,x_0,x_1,0)$$

and they are different. So, what is wrong in this proof?

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  • $\begingroup$ How exactly does Kosniowski define his standard simplices? $\endgroup$ – Lord Shark the Unknown May 7 '18 at 9:19
  • $\begingroup$ As a subspace of $\mathbb{R}^{n+1}$: $$\Delta_n = \{(x_0,\dots,x_n) \in \mathbb{R}^{n+1} : \sum_{i=0}^nx_i = 1, x_i \geq 0, i=0,1,\dots,n\}$$ $\endgroup$ – LH8 May 7 '18 at 10:11
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Note that $\delta_j\delta_i\phi$ takes an $(n-2)$-simplex. With $n=2$, it takes a $0$-simplex, not a $1$-simplex. Hence we have

$(\delta_1\delta_0\phi)(x_0)=(\delta_1(\delta_0\phi))(x_0)=(\delta_0\phi)(x_0,0)=\phi(0,x_0,0)$

Now, $\phi(0,x_0,0)=\delta_2\phi(0,x_0)=\delta_0\delta_2(x_0)$.

Personally, I find the definition $\delta_i$ a little bit confusing since it makes you think the composition from outside to inside, unlike the usual map composition. That might have induced you in error.

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    $\begingroup$ Me too. It is a little stranger. I still don´t see why is not explained that the composition works in inverse order. $\endgroup$ – LH8 May 7 '18 at 10:20
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    $\begingroup$ It's just the way it is defined. If $\phi$ is any $n$-singular simplex (it can also be the case that $\phi=\delta_j\varphi$ for some $(n+1$)-singular simplex $\varphi$), then $\delta_i\phi(x_0,\dots, x_{n-1})=\phi(x_0,\dots, x_{i-1},0,x_i,\dots, x_{n-1})$. So you have to follow this definition [...] $\endgroup$ – Javi May 7 '18 at 10:23
  • $\begingroup$ [...] if $\phi=\delta_j\varphi$, then you compute $\delta_j\varphi(x_0,\dots, x_{i-1},0,x_i,\dots, x_{n-1}))=\varphi(x_0,\dots, x_{i-1},0,x_i,\dots, x_{j-1},0,x_j,\dots, x_{n-1})$. $\endgroup$ – Javi May 7 '18 at 10:25
  • $\begingroup$ The reason is that $\delta_i\phi$ is defined by means of the arguments of $\phi$. Usually, a composition $f\circ g$ is defined as $f$ acting over the image of $g$, not over it's arguments (which are in its domain, not its image). $\endgroup$ – Javi May 7 '18 at 10:29

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