On a property of generating set for groups

Question

Consider the following paragraph from Lang's Algebra:

My question is about converse of one statement, which I was not able to prove.

Question: If $f(S)$ do not generates $F$, then can we find two homomorphisms $\psi_1,\psi_2$ which make diagram commutative in which we replace $\psi$ by $\psi_1$ and $\psi_2$?

Edited( after first answerby user218931): If ($S$ is non-empty and) $f(S)$ do not generate $F$, then does there exists a group $G$ with a map $g\colon S\rightarrow G$ for which there are two distinct $\psi$'s making diagram commutative?

Answer 1

In general, it will not be true that there exist two (distinct) homomorphisms $\psi_1$, $\psi_2$ making your diagram commute if $f(S)$ does not generate $F$.

Take for example $S = \mathbb Z$, $F = \mathbb Q$ and let $f\colon \mathbb Z\hookrightarrow \mathbb Q$ be the inclusion. Let $G$ be an abelian torsion-free group and $g\colon \mathbb Z\rightarrow G$ a homomorphism such that there exists a homomorphism $\psi\colon \mathbb Q\rightarrow G$ with $\psi\circ f = g$ ($G=\mathbb Q$ and $g=f$ provides a trivial example). Although $f(\mathbb Z) = \mathbb Z$ does not generate $\mathbb Q$ (as a group), $\psi$ is unique:
Let $\varphi\colon \mathbb Q\rightarrow G$ be another homomorphism with $f\circ \varphi = g$. Let $x\in \mathbb Q$ arbitrary and write $x = \frac nm$ with $n\in\mathbb Z$ and $m\in \mathbb N$. Since homomorphisms are $\mathbb Z$-linear, we obtain $$ m\cdot \left(\psi\left(\frac nm\right) - \varphi\left(\frac nm\right)\right) = m\cdot \psi\left(\frac nm\right) - m\cdot \varphi\left(\frac nm\right) = \psi(n) - \varphi(n) = 0, $$ where for the last equality, we have used that $\psi\big|_{\mathbb Z} = \varphi\big|_{\mathbb Z}$. Since $G$ is torsion-free, it follows that $\psi\left(\frac nm\right) - \varphi\left(\frac nm\right) = 0$, i. e. $\psi(x) = \varphi(x)$. Since $x\in \mathbb Q$ was arbitrary, we find $\psi = \varphi$ and hence $\psi$ is uniquely determined.

Edit: To answer the question in the comments, that given a set $S$, a group $F$ and a map $f\colon S\rightarrow F$ such that $f(S)$ does not generate $F$, there always exists a group $G$ together with a map $g\colon S\rightarrow G$ such that there exist two homomorphisms $\psi_1,\psi_2\colon F\rightarrow G$ with $\psi_i\circ f = g$ for $i=1,2$; this is true, at least in the case where $\langle f(S)\rangle$ is normal in $F$:
Take $G = \langle f(S)\rangle$ and $g\colon S\rightarrow G$, $s\mapsto 0$. Define $\psi_1\colon F\rightarrow G$ by $x\mapsto 0$ and let $\psi_2\colon F\rightarrow G$ by $x\mapsto \overline{x}$, the projection map. Obviously, $\psi_1\neq \psi_2$ (since $\langle f(S)\rangle \neq F$ by hypothesis) and we have $$ g(s) = \psi_1(f(s)) = 0 = \overline{f(s)} = \psi_2(f(s)) $$ for all $s\in S$. Therefore, such a group $G$ and a map $g\colon S\rightarrow G$ as above always exist.

Answer 2

There are certainly examples where the answer your question is yes. The simplest example is to take $S=\emptyset$ and any nontrivial group $F=G$. There is only one map from $S$ to $G$, so any homomorphism $\psi$ makes the diagram commute, and there are at least two (eg $\psi_1(x)=x$, $\psi_2(x)=1$).

On the other hand, for some choices of $F$ and $G$, $\psi$ may be uniquely determined regardless of $S$. For instance take $S=\emptyset$ again, and let $F=\mathbb Z_2$, $G=\mathbb Z_3$. There is only one homomorphism from $F$ to $G$.

Answer 3

Let $FS$ be the free group on the set $S$ and let $\bar f:FS\to F$ the the unique extension of your function $f$ to a group homomorphism. Since in the category of group epimorphisms are precisely the surjections and the map $\bar f$ is not surjective, we see that $\bar f$ is not an epimorphism. In other words, there exists a group $G$ and two morphisms $a,b:F\to G$ such that $a\neq b$ but $a\circ \bar f=b\circ\bar f$.

These $a$ and $b$ give you two $\psi$'s as in your question.