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Suppose $F_3(x,y,z)$ and $F_2(x,y)$ are free groups. Is it true that any map like $x\mapsto xy, y\mapsto z, z\mapsto y$ (or any other map which maps $x,y,z$ to reduced words in $F_2(x,y)$) define a group homomorphism? I'm confused because if I consider the image of $xy$ I cannot say whether it is the product of the images of $x$ and $y$ since the map above doesn't tell where the product $xy$ goes, it only specifies the images of single letters. Should I use the universal property, or this is something more obvious and basic?

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It's precisely the universal property: if $X$ is any set and $G$ any group, $f:X\to G$ an arbitrary mapping, the universal property of $F(X)$ (free group on $X$) yields a unique map $\overline{f}: F(X)\to G$ such that for $x\in X, \overline{f}(x) = f(x)$.

This is just the special case $X=\{x,y,z\}, G= F_2$.

In your example, $xy$ can only be sent to the product of $x$ and $y$, because $\overline{f}$ needs to be a morphism; and essentially this is how one can prove that the usual construction of the free group (with reduced words) has the correct universal property.

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