Boundary (operator) in singular homology

Question

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?

Answer 1

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.