# Right action induced by free group homomorphism

I was reading this proof by Steinberg of Nielsen-Schreier theorem and i had a doubt about proposition 1, that basically is an alternative universal property of free groups. It says:

Let $$X$$ be a set and $$F$$ a group equipped with a map $$i: X \rightarrow F$$. Then $$F$$ is a free group on $$X$$ (with respect to the mapping $$i$$) if and only if given any set $$A$$ and any map $$σ: X \rightarrow S_A$$, there is a unique action of $$F$$ on $$A$$ such that: $$ai(x)=\sigma(x)(a)$$

Note that in the paper when he says actions he will always refer to a right action. In the paper he says that the right implication was obvious and i thought so, but later when i tried to check it i came up with nothing. My idea was to use the universal property of free groups and then the inducted homomorphism:

$$F$$ is free so given the map $$\sigma$$ there's a unique homomorphism $$\varphi: F \rightarrow S_A$$ such that $$\varphi(i(x))=\sigma(x)$$ Then i thought that the unique actions could be the action $$*: A \times F \rightarrow A$$ with $$(a,w) \rightarrow \varphi(w)(a)$$. The only problem with that is that it is a left action and not a right action. To define the equivalent right actions that should be $$(a,w) \rightarrow \varphi(w^{-1})(a)$$ but this way the condition $$ai(x)=\sigma(x)(a)$$ is not true anymore.

Any ideas to prove this statement?

You just have to use the action $$(a,w)\to\varphi(w^{-1})(a)$$ for the unique homomorphism $$\varphi:F\to S_A$$ such that $$\varphi(i(x))=\sigma(x)^{\color{red}{-1}}$$.
Alternatively, I would guess that actually the author is using the reverse composition operation on $$S_A$$ (not the usual composition order of functions), so that a homomorphism to $$S_A$$ gives a right action rather than a left action.
• It is literally just an instance of the usual universal property, for the map $x\mapsto \sigma(x)^{-1}$. – Eric Wofsey Feb 3 at 2:26