0
$\begingroup$

Let $f:G \to H$ be a a group homomorphism such that for any two groups $H_1 , H_2$ , and any homomorphisms $g_1 : H \to H_1 , g_2 :H \to H_2$ , $g_1 \circ f = g_2 \circ f \implies g_1=g_2 $ ; then is it true that $f$ is surjective ?

$\endgroup$
3
$\begingroup$

Yes, it is a result of Schreier, but it is by no means elementary, see for instance this discussion of Arturo Magidin.

$\endgroup$
1
$\begingroup$

Yes this is true: consider $f$ is not surjective, i.e., $\forall g\in G, \exists h\in H$ such that $f(g) \neq h$. Then you can construct two group homomorphisms $g_1, g_2$ such that $g_1 \circ f = g_2 \circ f$ but $g_1 \neq g_2$ (which gives you your contradiction). I'll leave that construction up to you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy