Let $X$ and $Y$ be two sets with same cardinality. Let $\mathcal F (X)$ and $\mathcal F (Y)$ respectively denote the free groups generated by the sets $X$ and $Y.$ Can we say that $\mathcal F (X) \cong \mathcal F (Y)$ as groups?
I know that if $X$ is any non-empty set and $G$ is any group and $f : X \longrightarrow G$ be any map then there exists a unique surjective group homomorphism $\widetilde {f} : \mathcal F (X) \longrightarrow G$ such that $\widetilde {f} \circ j = f,$ where $j : X \longrightarrow \mathcal F (X)$ is defined by $j(x) = [x],\ x \in X.$
Can I use it anyway? Thanks.
Let $\phi\colon X\to Y$ be a bijection. Let $i\colon X\to \mathcal{F}(X)$ be the canonical embedding, and $j\colon Y\to \mathcal{F}(Y)$ be the canonical embedding.
The map $j\circ\phi\colon X\to \mathcal{F}(Y)$ induces a homomorphism $f\colon \mathcal{F}(X)\to \mathcal{F}(Y)$ that sends $x$ to $\varphi(x)$.
The map $i\circ\phi^{-1}\colon Y\to \mathcal{F}(X)$ induces a homomorphism $g\colon \mathcal{F}(Y)\to \mathcal{F}(X)$ that sends $y\to \phi^{-1}(y)$.
Now note that $g\circ f\colon \mathcal{F}(X)\to\mathcal{F}(X)$ sends $x$ to $x$ for all $x\in X$, and so must be the identity by the uniqueness clause of the universal property (it is induced by both $i$ and $g\circ f\circ i$; the former yields the identity, so we must have $\mathrm{id}=g\circ f$). Likewise, $g\circ f\colon \mathcal{F}(Y)\to \mathcal{F}(Y)$ sends $y$ to $y$ for all $y\in Y$, hence must be the identity (it is also induced by $g\circ f\circ j$). Thus, $f=g^{-1}$ are isomorphisms.