I want to use the universal property of the free group on two generators $F = F(\{a, b\})$ to construct infinitely many automorphisms of $F$, with the restriction that are not they are not inner automorphisms.
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1$\begingroup$ Indeed to understand the universal property, you need to have a clear idea of the meaning of "extend". If you have sets $X\subset Y$ and $Z$, and a map $f:X\to Z$, a map $g:Y\to Z$ extends $f$ if its restriction to $X$ equals $f$. Here this applies to $Y=F$ and $X=\{a,b\}$. $\endgroup$– YCorCommented Dec 1, 2019 at 10:05
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1$\begingroup$ If you want something concrete, consider the maps $a\mapsto a, b\mapsto ba^n$, $n\in {\mathbb Z}$. $\endgroup$– Moishe KohanCommented Dec 1, 2019 at 16:17
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$\begingroup$ Why is it not ok to define an arbitrary map, e.g. why is $a \mapsto a^n$, $b \mapsto b^n$ not ok? $\endgroup$– Tim HowellCommented Dec 3, 2019 at 5:39
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1$\begingroup$ @TimHowell Because it won't necessarily extend to an automorphism (that is an isomorphism/bijection). $\endgroup$– user29123Commented Dec 3, 2019 at 16:56
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It is well-known that $Aut(F_2)/Inn(F_2)\cong GL_2(\Bbb Z)$, see Bogopolski.
answered Dec 1, 2019 at 9:18