同伦类有限积结合性的归纳法证明验证及思路问询

Hey folks, I’ve been trying to prove this theorem about homotopy classes of paths, and I went with mathematical induction as my approach—but I’m not 100% sure if my reasoning holds up. Any hints or feedback on where I might adjust things, or how to strengthen this proof, would be massively appreciated!

Theorem #

Let $f_{1},f_{2},\dots,f_{n}$ be paths on a topological space $X$ such that $f_{i}(1)=f_{i+1}(0)$ for all $i=1,2,\dots,n-1$. Then:
$$[f_{1}]\ast[f_{2}]\ast\dots\ast[f_{n}]=[(((f_{1}\ast f_{2})\ast f_{3})\ast\dots)\ast f_{n}]$$

My Proof Attempt (by Induction on $n$) #

• Base Case ($n=2$):
  When $n=2$, the statement simplifies to proving $[f_{1}]\ast[f_{2}]=[f_{1}\ast f_{2}]$. This follows immediately from the definition of the $\ast$ operation on homotopy classes of paths.

• Inductive Hypothesis:
  Suppose the result holds for $n=k$, meaning:
  $$[f_{1}]\ast[f_{2}]\ast\dots\ast[f_{k}]=[(((f_{1}\ast f_{2})\ast f_{3})\ast\dots)\ast f_{k}]$$

• Inductive Step ($n=k+1$):
  We need to show the result holds for $n=k+1$. Let's break this down step by step:
  $$
  \begin{align*}
  [f_{1}]\ast[f_{2}]\ast\dots\ast[f_{k+1}]&=([f_{1}]\ast[f_{2}]\ast\dots\ast[f_{k}])\ast[f_{k+1}] \
  &=[(((f_{1}\ast f_{2})\ast f_{3})\ast\dots)\ast f_{k}]\ast[f_{k+1}] \
  &=[((((f_{1}\ast f_{2})\ast f_{3})\ast\dots)\ast f_{k})\ast f_{k+1}]\
  &=[(((f_{1}\ast f_{2})\ast f_{3})\ast\dots)\ast f_{k+1}].
  \end{align*}
  $$

That's where I've gotten so far—does this induction step check out? Am I missing any key details that would make the proof more rigorous?

备注：内容来源于stack exchange，提问作者PHR