I have been staring at the definition for days, drew diagrams but I don't understand as to what its elements are
The fundamental group $\pi_1(X,x)$ at a base point $x$ is a set of rel $\{0,1\}$ homotopy classes $[\alpha]$ of closed paths $\alpha: I \to X$ such that
$$\alpha(0)=\alpha(1)=x \in X$$
with concatenation of paths as its operation.
Now, looking at this, I drew a diagram. Draw some $X$ and a point $x$ in it. The condition
$$\alpha(0)=\alpha(1)=x \in X$$
tells me I am looking at only paths that start from $x$ and end at $x$. i.e. A loop. Of course, I can draw various and various loops. One that looks like an oval, one that could go all over $X$ and come back to $x$, or the constant path.
By the homotopy rel $\{0,1\}$ condition, I consider some homotopy $h$ between paths(loops) that give
$$h(0,t)=\alpha(0)=\beta(0) \text{ and } h(1,t)=\alpha(1)=\beta(1)$$
For any loops $\alpha,\beta$ with base point $x$.
Now I must admit; I am very confused with the amount of fancy words here, homotopies, rel $B$, paths, base point, homotopy classes etc.
But looking at the above, I thought "doesn't a rel $\{0,1\}$ homotopy exist between any two loops in $X$ about $x$?" Namely, this leads to the (wrong) conclusion that there is only one element in the fundamental group since homotopy is an equivalence relation and we are considering its equivalence classes(homotopy class).
Because,
Any loop about $x$ in $X$ have the property $\alpha(0)=\alpha(1)=x \in X$. So these types of paths are the ones we consider. By definition of a homotopy between some $\alpha, \beta$,
$$h(s,0)=\alpha(s) \text{ and } h(s,1)=\beta(s)$$
For any two given paths, let me define $h(s,t)=(1-t)\alpha(s)+t\beta(s)$. Then,
$$h(s,0)=\alpha(s), h(s,1)=\beta(s)$$
Thus a homotopy between $\alpha,\beta$. Furthermore, as mentioned earlier we only consider loops based at $x$ so we have for both $\alpha,\beta$
$$h(0,t)=(1-t)\alpha(0)+t\beta(0)=\alpha(0)-t\alpha(0)+t\beta(0)$$
Since $\alpha(0)=\alpha(1)=x=\beta(0)=\beta(1)$ as a loop,
$$\alpha(0)-t\alpha(0)+t\beta(0)=\alpha(0)=\beta(0)$$
We can compute $h(1,t)$ in the exact same way to obtain $\alpha(1)=\beta(1)=h(1,t)$, namely giving us $h(s,t)$ as the homotopy rel $\{0,1\}$. Then, since $\alpha,\beta$ were arbitrary loops we can define such homotopy for any two loops about the base point $x$. Therefore, by transitivity of homotopy as equivalence relations, I have that every loop is homotopic rel $\{0,1\}$ to each other.
$\alpha \sim \beta \Leftrightarrow \exists h(s,t):\alpha \cong \beta$ such that $h$ is homotopy rel $\{0,1\}$.
Namely, every loop is homotpic rel $\{0,1\}$ to each thereby we only ave one equivalence class. i.e. the elements of the fundamental group turns out to be only one.
So that's my reasoning. Of course I know it's wrong, but I don't know why it's wrong. A misunderstanding of definitions? Abuse of definitions? Abuse of logic? What drove me in this wrong direction?
I really need help on this