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