How to show that a homomorphism between fundamental groups is an inner automorphism?

Question

I want to show that if $\sigma$ is a loop based at a point $p \in X$, where $X$ is a topological space, then the homomorphism $\Phi _\sigma: \Pi_1(X,p) \rightarrow \Pi_1(X,p)$, is an inner automorphism.

The general homomorphism between the fundamental groups that I've been using is $\Phi _\gamma: \Pi_1(X,p) \rightarrow \Pi_1(X,q)$ where $\Phi_\gamma$ maps the equivalence class $[\alpha] \in \Pi_1(X,p) \mapsto [\gamma^{-1}\alpha\gamma]$.

Thus, $\Phi_\sigma$ would map $[\alpha] \mapsto [\sigma^{-1}\alpha\sigma]$. However, in order to show that $\Phi_\sigma$ is an inner automorphism, I know I need to show that the homomorphism actually maps $[\alpha] \mapsto [\sigma][\alpha][\sigma]^{-1}$, in addition to satisfying the other properties of automorphisms.

I know instinctually that I can separate the equivalance class $[\sigma \alpha \sigma ^{-1}]$ because $[\sigma] \in \Pi_1(X,p)$. However, I'm not sure as to how to show that rigorously.

I'd appreciate any guidance in doing so.

Edit: Thank you Nishant for helping me realize that the equivalence classes can be separated. That was really quite obvious. I noticed, however, that I wrote my original homomorphism incorrectly. It has now been corrected.

As a result, I now need to figure out how to get $ [\sigma^{-1}\alpha\sigma] $ to $[\sigma][\alpha][\sigma]^{-1}$. Is it because $[\sigma^{-1}]=[\sigma]$? If so, how would I show that? Otherwise, am I completely off base in how I am approaching this question?

Answer 1

You're really close. As you say, you now know that $[\sigma^{-1}\alpha\sigma]=[\sigma^{-1}][\alpha][\sigma]=[\sigma]^{-1}[\alpha][\sigma]$ and so now you just need to let $g=[\sigma]^{-1}$ so that $g^{-1}=([\sigma]^{-1})^{-1}=[\sigma]$, hence $[\sigma^{-1}\alpha\sigma]=g[\alpha]g^{-1}$. So the automorphism $\Phi_{\sigma}$ maps $[\alpha]\mapsto g[\alpha]g^{-1}$ which is an inner automorphism.

In practice, because we can always just replace a conjugating element by its inverse, we say that both $gag^{-1}$ and $g^{-1}ag$ are conjugates of $a$ (though by different elements, and dependent on your convention of acting on the left or right).