I have difficulties with this task in linear algebra:
By mathematical induction, prove: let any finite mapping $f_1$,$f_2$,$\cdots$,$f_n$ exist. If for any $1\leqslant i\leqslant n-1$, the domain of $f_{i+1}$ equals the codomain of $f_i$, the composition of functions is always generalized associative. That is to say, the result of $f_n\circ\cdots\circ f_2\circ f_1$ is independent of how the expression is bracketed.
For example: $$((f_4\circ f_3)\circ f_2)\circ f_1=(f_4\circ f_3)\circ(f_2\circ f_1)=(f_4\circ(f_3\circ f_2))\circ f_1=f_4\circ((f_3\circ f_2)\circ f_1)=f_4\circ(f_3\circ(f_2\circ f_1)).$$
It is already proved that for $n=3$, the statement holds. My attempt is as follows.
Assume for any given case $3\leqslant n\leqslant k (n\in\mathbb{N}^*,k\geqslant3,k\in\mathbb{N}^*)$ the statement holds. Then while $n=k+1$, $$f_{k+1}\circ\cdots\circ f_2\circ f_1=(f_{k+1}\circ\cdots)\circ(f_{i+1}\circ f_{i})\circ(\cdots\circ f_2\circ f_1)=\underbrace{(f_{k+1}\circ\cdots)\circ(g)\circ(\cdots\circ f_2\circ f_1)}_{k\text{ terms}},$$ and $1\leqslant i\leqslant k$, $i\in\mathbb{N}^*$, so the statement applies too.
I want to know if:
- There are any corrections to be made, and
- Better wording for the proof.
Thank you all in advance.
