Simple exercise in fundamental groups and identification degree I know this is simple, but I'm not being able to do this.
a) Let $\Phi : \mathbb{S}^1 \times \mathbb{S}^1 \to \mathbb{S}^1$ be the multiplication of complex numbers, $\Phi (z_1, z_2) = z_1z_2$ and let $\omega_j: \mathbb{I} \to \mathbb{S}^1$ be any two loops based at 1. Define $(ω_1 • ω_2)(t) = \Phi ◦ (ω_1(t), ω_2(t))$
i) Show that $[ω_1 • ω_2] = [ω_1 ∗ ω_2]$.
ii) Under the identification deg : $π_1(\mathbb{S}^1) \cong \mathbb{Z}$, show that $Φ_\# : π_1(\mathbb{S}^1 × \mathbb{S}^1) → π_1(\mathbb{S}^1)$ is given by $Φ_\#(a, b) = a + b$.
 A: Let $(\Bbb S^1,\times)$ be the topological group with identity element $1$, where $\times$ denotes the usual multiplication of complex numbers. For any two continuous functions $\alpha,\beta:[0,1]\to \Bbb S^1$, define $\alpha\bullet\beta:[0,1]\to \Bbb S^1$ as $\alpha\bullet\beta(t)=\alpha(t)\times\beta(t)$ for all $t\in [0,1]$. Let $C_1$ be the constant loop in $\Bbb S^1$ based at $1$. Also, $*$ stands for concatenation of two paths.

For two loops $\gamma,\delta$ in $\Bbb S^1$ based at $1$ we have $\gamma*C_1\simeq_{\text{rel }1}\gamma\simeq_{\text{rel }1} C_1*\gamma$ as $C_1$ represents the identity element of $\pi_1(\Bbb S^1,1)$, so that  $$\gamma*\delta=(\gamma*C_1)\bullet (C_1*\delta)\simeq_{\text{rel }1}\gamma\bullet \delta\simeq_{\text{rel }1}(C_1*\gamma)\bullet  (\delta*C_1)=\delta*\gamma.$$
In the above, to check the "$=$" just plugin an arbitrary $t\in [0,1]$ both left and right-hand sides. So, we completed part $(i)$.

Note that $\deg:\pi_1(\Bbb S^1,1)\to \Bbb Z$ is a group isomorphism.
Also, $\pi_1\big(\Bbb S^1\times \Bbb S^1,(1,1)\big)\cong\pi_1(\Bbb S^1,1)\times \pi_1(\Bbb S^1,1)$ via projections induced maps. So, write an element of $\pi_1\big(\Bbb S^1\times \Bbb S^1,(1,1)\big)$ as $\big([\omega_1],[\omega_2]\big)$ for some $[\omega_1],[\omega_2]\in \pi_1(\Bbb S^1,1)$. Then, $$\Phi_\#\big([\omega_1],[\omega_2]\big)=[\omega_1\bullet\omega_2]=[\omega_1*\omega_2].$$ So, using degree isomorphism $$\Phi_\#\big(\deg[\omega_1],\deg[\omega_2]\big)=\deg[\omega_1*\omega_2]=\deg[\omega_1]+\deg[\omega_2].$$
