Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ and $B$ be non-empty sets with associated free groups $F(A),F(B)$. Given an injective function $f: A \to B$, is the induced homomorphism $\bar{f}: F(A)\to F(B)$ injective?

Let $i_A: A \to F(A)$ and $i_B: B \to F(B)$ be the usual inclusions. Taking $i_A$ and $i_B \circ f$ we get an induced homomorphism $\bar{f}: F(A) \to F(B)$, but it cannot be concluded that $\bar{f}$ is injective using only the universal property of free objects.

Given two distinct words $w,w' \in F(A)$, the claim should follow by contradiction if we can conclude that $\bar{f}(w)=\bar{f}(w')$ implies $\bar{f}(a_1) = \bar{f}(a_2)$ for some distinct $a_1,a_2 \in i_A(A)$. Is this conclusion possible? For context I am looking at the construction of the free product given here.

On a side note, the construction of the injections in the linked document seems slightly ambigious.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Yes, $\bar f$ is injective. To prove it, use the universal property of $F(B)$ to define a homomorphism $h:F(B)\to F(A)$ by defining it on the generators in $B$ as follows. For generators $b$ in the range of $f$, say $b=f(a)$, set $h(b)=a$. (That's unambiguous because $f:A\to B$ is one-to-one.) For generators $b$ not in the range of $f$, set $h(b)=e$, the identity element of $F(A)$. Notice that $h\circ\bar f$ is a homomorphism from $F(A)$ into itself, sending each generator $a\in A$ to itself. But the identity map $1$ of $F(A)$ is also such a homomorphism. By the uniqueness clause in the universal property of $F(A)$, it follows that $h\circ\bar f=1$, and that implies in particular that $\bar f$ is one-to-one.

share|improve this answer
1  
The same proof works in every algebraic category. –  Martin Brandenburg Sep 8 '13 at 14:23
    
@MartinBrandenburg Right as long as $F(\varnothing)$ is nonempty (the proof needs something to play the role of the "garbage" value $e$). Meanwhile, here's a quicker proof of the result whenever $A$ isn't empty: In the category of sets, every monomorphism with nonempty domain is split, and all functors preserve split monos. –  Andreas Blass Sep 8 '13 at 21:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.