Why the fundamental group of the circle is not trivial?

Question

What is wrong with my reasoning?

1. $S^1$ is path connected
2. Let $x_0\in S^1$ and $[f]\in \pi_1(S^1,x_0)$.
3. $f$ is a loop in $x_0$, i.e, $f:[0,1]\to S^1$ with $f(0)=f(1)=x_0$.
4. for any $s\in [0,1]$, exists $\alpha_s:[0,1]\to S^1$ path with $\alpha_s(0)=f(s)$ and $\alpha_s(1)=x_0$ (because $S^1$ is path connected)
5. Let $H:[0,1]\times [0,1]\to S^1$, $H(t,s)=\alpha_{s}(t)$. Then $H(0,s)=\alpha_{s}(0)=f(s)$ and $H(1,s)=\alpha_{s}(1)=x_0$.
6. Therefore $H$ is homotopy, then $f$ homotopic to $e_{x_0}$ (trivial loop in $x_0$). Therefore $[f]=[e_{x_0}]$
7. Therefore $\pi_1(S^1,x_0)=\left\{[e_{x_0}]\right\}$... but, it knowing that $\pi_1(S^1,x_0)=\mathbb{Z}$ for any $x_0\in S^1$.

Answer 1

In the definition of homotopy group, the homotopies deforming paths must fix the basepoint in 0 and 1. You should therefore have that $\alpha_s(0) = x_0= \alpha_s(1) $.

Also, from your construction it's not evident that $\alpha_s$ can be chosen continuously in $s$!