Universal Mapping Property of Free Groups: Definition and Proof

[fishturtle1]

Homework Statement

I'm confused about my notes about free groups, looking for help to understanding them.

Relevant Equations

will put definitions below

Let $S = \lbrace a, b \rbrace$ and define $F_S$ to be the free group, i.e. the set of reduced words of $\lbrace a, b \rbrace$ with the operation concatenation. We then have the universal mapping property: Let $\phi : S \rightarrow F_S$ defined as $s \mapsto s$ and suppose $\theta : S \rightarrow G$ is any function where $G$ is a group. Then there exists unique homomorphism $f : F_S \rightarrow G$ such that $\theta = f \circ \phi$. For example, $f(aba^{-1}) = f(a)(f(b)f(a)^{-1} = \theta(a)\theta(b)\theta(a)^{-1}$.

My question is, where does $f$ come from? It just seems like there is some step that I am missing to get $f$ in the first place?

 

[member 587159]

This theorem just says that any map $\theta$ assigning letters to elements of a group $G$, extends uniquely to a morphism $f$ on the free group on these letters to $G$.

This is what one uses when one says: 'A homomorphism of groups is determined completely if you specify the images of the generators' (and in the image the relations that the generators satisfy must also be satisfied, but this is superfluous here because there are no relations between the letters that generate the free group).

So, to be concrete. Consider any group $G$. Fix two elements $g,h \in G$ and consider the map $\theta$ by

$$a \mapsto g, b \mapsto h$$

The theorem says that there is a unique group morphism

$$F_S = \langle a ,b \rangle \to G$$

satisfying $a \mapsto g, b \mapsto h$.

 

[fishturtle1]

This theorem just says that any map $\theta$ assigning letters to elements of a group $G$, extends uniquely to a morphism $f$ on the free group on these letters to $G$.

This is what one uses when one says: 'A homomorphism of groups is determined completely if you specify the images of the generators' (and in the image the relations that the generators satisfy must also be satisfied, but this is superfluous here because there are no relations between the letters that generate the free group).

Ok thank you. And so $f$ is defined to be a homomorphism from the beginning?

 

[member 587159]

Ok thank you. And so $f$ is defined to be a homomorphism from the beginning?

I edited my previous post to make it more concrete. But the theorem says that $f$ must be a homomorphism. The map $\theta$ does NOT have be a homomorphism.

 

[fishturtle1]

I edited my previous post to make it more concrete. But the theorem says that $f$ must be a homomorphism. The map $\theta$ does NOT have be a homomorphism.

I think that makes sense, I'm going to try some problems.