Consider the following proposition $13.14$:
I'm confused with the fourth line in the proof. The author says "we claim that there exist $t_1 \in (t_0,1]$ such that $(P_1 \circ \gamma)(t_0,t_1) \subseteq K$"
Why we claim this? What is the necessity to proving this claim? Is there anything special for this claim?
Can anybody explain it a bit more?
Still one more question: How $(P_1 \circ \gamma)(t_0)=0$ in the last line?
Added: Here's the more detailed expansion of the deleted comb space and its figure:
Here $D$ is same as $C_0$ in our proposition
Best Answer
I will answer your questions in a second, but here is what's going on in the proof. If there were a continuous path from $0\times 1$ to $1\times 0$, then the path must "leave" the point $0\times 1$ at some point $t_0$. The question becomes: where does the path go right after $t_0$? Well, it can't really go anywhere, and that's the whole point.
Your first question of "why do we prove this claim"? You would know the answer to this question if you understood the rest of the proof, so I'll try explaining the rest of the proof.
Look at the set $(P_1\circ \gamma)(t_0,t_1)$. According to the claim, it is a subset of $K$. Since $(t_0,t_1)$ is a connected interval and $P_1\circ\gamma$ is a continuous function, $(P_1\circ \gamma)(t_0,t_1)$ is a connected subset of $K$. But then it must be a singleton. This is a huge deal. What it's telling us is that $\gamma(t)$ has the same $x$-coordinate for $t \in (t_0,t_1)$, i.e. that right after we "leave" the point $0\times 1$, we get stuck at some vertical line with $x$-coordinate some point of $K$. And this is of course a contradiction since at $t_0$, $\gamma$ is at $0\times 1$, so $\gamma$ clearly cannot be continuous. [We took $t_0$ to be the maximum of $\gamma^{-1}(\{p\})$ because this is the first time $\gamma$ "leaves" $0\times 1$].
Finally, "How is $(P_1\circ \gamma)(t_0) = 0$?". $\gamma(t_0) = 0\times 1$ by definition of $t_0$. So, $(P_1\circ \gamma)(t_0) = 0$.