Proof explanation, prop 7.16 "Introduction to Topological Manifold" by Lee I have a question about the proof of proposition 7.16 in Lee's Topological Manifold text.  The proof is shown below. 

My first question is that for the first paragraph when he shows a) implies b), the map $ \tilde{H} $ is a homotopy that fix the endpoint $p$ for all $t \in I$ , is that correct,  since $\tilde{H}( (1,0), t) = p $ for all $t \in I$ due to the fact that $f$ is a loop based at $p$?  So then $ \tilde{f}$ is not freely homotopic to a constant map then ??
My second question is about the part where he shows b) implies c).  I don't get the last sentence of that paragraph.  what does he mean by "because the quotient map restricts to the obvious identification $~~ \mathbb{S}^1 \times \{1\} \approx \mathbb{S}^1 \hookrightarrow \overline{\mathbb{B}}^2$ ??
Thank you. 
 A: For your second question, we start with our map $H:\Bbb S^1\times I\to X$, and by definition of $H_0$ being a constant map, all of $\Bbb S^1\times\{0\}$ is sent to the same point and therefore we get an induced map $\tilde H:C\Bbb S^1\to X$, which is the same as a map $\tilde H:\bar{\Bbb{B}}^2\to X$.
Let $\pi:\Bbb S^1\times I\to (\Bbb S^1\times I)/(\Bbb S^1\times\{0\})=C\Bbb S^1\cong\bar{\Bbb{B}}^2$ be our quotient map. Do you have an image of what the latter homeomorphism looks like? Imagine the space $C\Bbb S^1$, which is just a cone:

The homeomorphism is obtained by "pushing everything down". When we do this, where does $\Bbb S^1\times\{1\}$ go? It just gets sent to the unit circle.
So when we restrict $\tilde H$ to $\Bbb S^1\times\{1\}$, we're really restricting it to the unit circle. But this restriction is also equal to the restriction of $H$ to $S^1\times\{1\}$, which by definition is $\tilde f$. Hence $\tilde H$ is a map $\bar{\Bbb{B}}^2\to X$ such that $\tilde H|_{\Bbb S^1}=\tilde f$, as desired.
