Proof in Lee's Introduction to Topological manifolds

Question

In his book, Lee has what appears to be a very simple proof of the fact that if $f:I\to X$ is a path in $X$, then $f^{-1}\sim -f$ where $f^{-1}(t)=f(1-t).$ He takes $\Delta_2$ to be the simplex with vertices $(0,0), (1,0), (0,1)$ and defines $\sigma:\Delta_2\to X$ by $\sigma (x,y)=f(x)$. Let $c_q$ be the constant map at $f(0)=q.\ $ Of course, $c_q$ is a boundary. Now, he claims that $\partial \sigma=f^{-1}-c_q+f,\ $ from which the result obtains. But I can't seem to get the first term, that is, $f^{-1}.$ Since $\partial \sigma$ is a formal alternating sum of the restrictions of $f$ to the faces. So, wouldn't we have simply $\partial \sigma =f-c_q+f?$

I am not used to working with simplices described in this way, so I tried to work the problem out by taking $\Delta_2$ to be the "usual" simplex, with vertices $e_o=(1,0,0);\ e_1=(0,1,0);\ e_2=(0,0,1).$ Defining $\sigma$ as Lee does, using these data, I still do not get his result.

I am curious to see where the problem is, because the proofs I have seen of this fact, while not terribly difficult, do require some work, while Lee's is almost trivial.

Answer 1

The simplex $(0,0); (1,0); (0,1)$ is a right triangle. The hypothenuse that is the segment $(1,0);(0,1)$ is parameterized by $c(t)=(1-t,t)$ so $\sigma(c(t))=f^{-1}(t)$.

Answer 2

It's a simple calculation, and you just have to be careful. Following Lee's diagram, set $e_0=(0,0);\ e_1=(1,0);\ e_2=(0,1)$.

Then $\sigma (re_0+se_1+te_2)=\sigma (se_1+te_2)=f(s)$ by definition of $\sigma.$

Now,

$\partial \sigma(r,s)=\sigma (0,r,s)-\sigma (r,0,s)+\sigma (r,s,0)=$

$\sigma (0e_0+re_1+se_2)-\sigma (re_0+0e_1+se_2)+\sigma (re_0+se_1+0e_2)=$

$f(r)-f(0)+f(s)=f(r)-q+f(1-r).$

Thus, $\partial \sigma=f-c_q+f^{-1},\ $ as desired.