Let $G$ be a topological group with neutral element $e$ and let $I:G\to G$ denote the inversion. Is it true that if $[\gamma]\in\pi_{1}(G,e)$, then $[I\circ\gamma]=[\overline{\gamma}]$, where $\overline{\gamma}$ denotes the inverse path, i.e. $\overline{\gamma}(t):=\gamma (1-t)$?
I know that there is a similar statement for the multiplication $\mu:G\times G\to G$, namely that $\mu_{\ast}([\gamma_{1}],[\gamma_{2}])=[\gamma_{1}\cdot\gamma_{2}]$, where $\cdot$ denotes the concatenation.
Let $(G,\bullet)$ be a topological group with identity element $e$. For any two continuous functions $\alpha,\beta:[0,1]\to G$, define $\alpha\bullet\beta:[0,1]\to G$ as $\alpha\bullet\beta(t)=\alpha(t)\bullet\beta(t)$ for all $t\in [0,1]$. Similarly define, $\alpha^{-1}:[0,1]\to G$ as $\alpha^{-1}(t):=\big(\alpha(t)\big)^{-1}$ for all $t\in [0,1]$. Let $C_e$ be the constant the loop in $G$ based at $e$. Also, $*$ stands for concatenation of two paths.
For two loops $\gamma,\delta$ in $G$ based at $e$ we have $\gamma*C_e\simeq_{\text{rel }e}\gamma\simeq_{\text{rel }e} C_e*\gamma$ as $C_e$ represents the identity element of $\pi_1(G,e)$, so that $$\gamma*\delta=(\gamma*C_e)\bullet (C_e*\delta)\simeq_{\text{rel }e}\gamma\bullet\delta\simeq_{\text{rel }e}(C_e*\gamma)\bullet (\delta*C_e)=\delta*\gamma$$$$\text{ and putting }\delta=\gamma^{-1}\text{ we have, }$$$$\gamma*\gamma^{-1}\simeq_{\text{rel }e}\gamma\bullet \gamma^{-1}= C_e=\gamma^{-1}\bullet \gamma\simeq_{\text{rel }e}\gamma^{-1}*\gamma$$$$\implies \gamma^{-1}\simeq_{\text{rel }e}\overline \gamma$$$$\text{ as }\overline\gamma \text{ represents the inverse of }[\gamma]\in \pi_1(G,e) .$$
In the above, to check the "$=$" just plugin an arbitrary $t\in [0,1]$ both left and right-hand sides.