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Let $S,T$ be sets such that $F(S)\subset F(T)$ ($F(S),F(T)$ are the free groups of $S,T$)

Assume $S\subset T$.

Since $F(S)$ is free, there exists a unique homomorphism $\phi:F(S)\rightarrow F(T)$ such that $\phi(x)=x$ on $S$.

How do I prove that $\phi$ is injective?

Rubertos
  • 12,491

2 Answers2

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Let $f \colon T \to S$ be any mapping that restricts to the identity on $S$. Then $f$ induces a morphism $\psi \colon F(T) \to F(S)$ such that $\psi \circ \phi = \operatorname{id}_S$.

user180040
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Below I write $\circ$ for the group operation.

Let's assume that $\phi(x) = \phi(y)$ where $x = \langle s_1, ..., s_m \rangle$ and $y = \langle t_1, ..., t_n \rangle$ and each $s_i$ as well as $t_j$ is in $S$. Since $\phi$ is a homomorphism, we know that $\phi(x) = \phi(s_1) \circ ... \circ \phi(s_m)$. But by construction $\phi(s_i) = s_i$, hence $\phi(x) = x$ and likewise $\phi(y) = y$. That means $x = y$ and $\phi$ is injective.

Martin Berger
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