TITLE: A confusion about the fact that contractible spaces are simply connected
QUESTION [6 upvotes]: Question 1: Greenberg's Algebraic topology has a proof that contractible spaces are simply connected. In the middle of the proof, the book makes use of the following fact without justifying it (probably because the author thinks its useful):

Fact 1: Let $p:\mathbb{I}\rightarrow X$ be a continuous function such that
  $p(0)=p(1)=x_0$ for some $x_0\in X$ . Let $c:\mathbb{I}\rightarrow X$ send every element of the interval to $x_0$. If $p,c$ are homotopic, then there exists a continuous
  function $F:\mathbb{I}\times \mathbb{I}\rightarrow X$ such that:
1) $F(s,0)=c(s)$ for all $s\in \mathbb{I}$
2) $F(s,1)=p(s)$ for all $s\in \mathbb{I}$
3) $F(0,t)=F(1,t)$ for all $t\in \mathbb{I}$

Since the book is assuming this fact as obvious, I suppose it should be much easier to prove than to prove that contractible spaces are simply connected. I asked this here before and I got comments implying that fact 1 is almost as hard as proving the fact that "contractible spaces are simply connected". Well if this were true, is the proof of Greenberg's algebraic topology bad because it is assuming that fact 1 is obvious when proving that contractible spaces are simply connected while fact 1 is as hard as proving the original claim "contractible spaces are simply connected" ????
Important remark: I noticed that when I asked this question before, people were trying to prove the fact that "contractible spaces are simply connected" instead of answering the question. I know that it is tempting to answer this question without reading it carefully and just proving the fact that "contractible spaces are simply connected", please try to avoid this temptation :) 
Question 2: Let's assume that fact 1 is true, since it's mentioned in the book.  The following fact is also mentioned in the book. I will state it without proof:

Fact 2: Let $F:\mathbb{I}\times\mathbb{I}\rightarrow X$ be continuous such that $F(0,0)=F(0,1)=F(1,0)=F(1,1)$. Let $\gamma,\delta,\alpha,\beta :\mathbb{I}\rightarrow X$
  be given by $\gamma
(s)=F(s,0),\delta(s)=F(s,1),\alpha(t)=F(0,t),\beta(t)=F(1,t)$, then
  $\delta $ is homotopic to $\alpha^{-1} \gamma \beta$ $rel\{0,1\}$

Claim: Let $p:\mathbb{I}\rightarrow X$ be continuous such that $p(0)=p(1)=x_0$ for some $x_0\in X$. Let $c:\mathbb{I}\rightarrow X$ send every element of the interval to $x_0$. If $p,c$ are homotopic, then $p,c$ are homotopic rel$\{0,1\}$
Proof: Since $p,c$ are homotopic, therefore fact 1 can be used to show that there exists a homotopy $F$ between $p,c$ such that $F(0,t)=F(1,t), F(s,0)=c(s),F(s,1)=p(s)$ for all $t\in\mathbb{I}$. Set $\alpha(t)=F(0,t),\beta(t)=F(1,t)$. From fact 1, we know that $\alpha=\beta$. From Fact 2, we know that $p$ is homotopic to $\alpha^{-1} c \alpha$ rel $\{0,1\}$. Since $\alpha^{-1}c\alpha$ is homotopic  to $\alpha^{-1}\alpha$ rel $\{0,1\}$ (because $c$ is constant map) , the last is homotopic to $c$ rel $\{0,1\}$. Therefore $p,c$ are homotopic rel $\{0,1\}$.
What is wrong with my argument in proving the above claim ?

I will explain here why I think my claim is not correct:
Consider the space $S$ formed as the identification space that results from gluing the the subspace $\mathbb{I}\times\{0\}\cup\{(0,1),(1,1)\}$  of $\mathbb{I}^2$ to one point. Let $F:\mathbb{I^2}\rightarrow S$ be the identification map. The space $S$ looks like the space resulting from gluing the three vertices of a triangular sheet of paper together (I already tried it using paper). Now note that $F:\mathbb{I}\times \mathbb{I}\rightarrow X$ is a homotopy between the paths $F|\mathbb{I}\times \{0\}$ and $ F|\mathbb{I}\times \{1\}$. I fail to see that the paths: $F|\mathbb{I}\times \{0\},F|\mathbb{I}\times\{1\}$ are homotopic rel $\{0,1\}$ visually on the space $S$ that I created using paper. Thus, this is a potential counterexample to the claim .
If you have time, could you please construct the space $S$ from paper just like I did to understand me better ?

Question 3: I want to see a proof that contractible spaces are simply connected using the knowledge/definitions I have so far. I only know the definition of homotopic, homotopic rel $\{0,1\}$. Simply connected means having a trivial fundamental group. Contractible means that the identity map $1_X:X\rightarrow X$ is homotopic to some constant map $c:X\rightarrow X$
Finally thank you for wasting your time in answering my (probably) trivial questions

REPLY [3 votes]: Question 1: I apologize again for trying to prove Fact 1. It is false, as Ryan already explained. A correct similar statement is that if $X$ is contractible, then for any path $p$ there exists a homotopy $F$ satisfying those conditions.
Question 2: Claims 2 is also false, and your proof of the claim is correct except for using Fact 1 which is false. By changing the first sentence of your proof you can turn it into a correct proof of the correct claim that if $X$ is contractible, for any loop $p$, $p$ and the constant map $c \equiv p(0) = p(1)$ are homotopic relative to endpoints.
Question 3: Here is how I think of the proof that contractible implies simply-connected. The whole thing boils down to showing your Fact 2: the definition of contractible you're using only gives homotopies that are not relative to endpoints, Fact 2 is the thing that converts those into homotopies rel endpoints.
I will use the following notation and facts:

By $p \sim q$ I'll mean that the two paths $p, q : [0,1] \to X$ are homotopic relative to the endpoints. (I won't use any symbol for homotopies that are not between paths or are not relative to endpoints.)
I'll use $q \cdot p$ to mean the concatenation of paths $p$ and $q$ (doing $p$ first and then $q$, say), and I'll need to use that this operation

respects homotopy in the sense that $p_1 \sim p_2 \implies q \cdot p_1 \sim q \cdot p_2$,
is homotopy associative, $p \cdot( q \cdot r) \sim (p \cdot q) \cdot r$, and
has homotopy units given by constant paths $c_x$, i.e., $p \cdot c_{p(0)} \sim p$, $c_{p(1)} \cdot p \sim p$.
has homotopy inverses given by traversing paths in reverse order, $p \cdot p^{\leftarrow} \sim c_{p(1)}$, $p^{\leftarrow}  \cdot p \sim c_{p(0)}$.


I'm sure you've developed all of that if you know what the fundamental group is.
Now, let $X$ be contractible and choose a homotopy $H : X \times [0,1] \to X$ between the identity and a constant path $c_{x_0}$. Each point $x \in X$ then gets a chosen path $h_x$ connecting it to $x_0$, given by $h_x (t) = H(x,t)$. We can use your Fact 2 to produce some a homotopy relative endpoints for any path $p$: applying Fact 2 to the homotopy $(s,t) \mapsto H(p(s),t)$ we get a homotopy showing that $h_{p(1)} \cdot p \cdot h_{p(0)} ^ {\leftarrow} \sim c_{x_0}$. Using the above facts for the operation $\cdot$, we get that $p \sim h_{p(1)} ^\leftarrow \cdot h_{p(0)}$. This shows that all paths between $p(0)$ and $p(1)$ are homotopic to each other relative to endpoints because they are all homotopic to $h_{p(1)} ^\leftarrow \cdot h_{p(0)}$. This proves in particular that $X$ is simply-connected, because if $p(0)=p(1)$ we get $p \sim h_{p(1)} ^\leftarrow \cdot h_{p(0)} \sim c_{p(0)}$.