Map is surjective/injective and a short exact sequence of groups Im working on my thesis about semidirect products and splitting lemma. I got the following theorems to prove and Im a not sure how to start. I would appreciate any help.
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1. Let $f:A\to B$ be a map.
Show:
a) if $g:B\to A$ so that $gf=id_{A}$ then $f$ is injective
b) if $g:B\to A$ so that $fg=id_{B}$ then $f$ is surjective
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*$A$, $B$, $G$ groups and there is a short exact sequence
$1\to A\to G\to B\to 1$
then $\alpha :A\to G$ is injective and $\beta :G\to B$ is surjective. Please show that.
 A: Finding the proof is practically unavoidable as soon as you make yourself clear what you are given and what you want to show:
1a) Assume $f(y)=f(y)$. You want to show that this implies $x=y$. You are given the fact that $g(f(t))=t$ for all $t\in A$. Aha: $f(x)=f(y)$ imlpies $x=g(f(x))=g(f(y))=y$.
1b) Assume $b \in B$. You want to exhibit some $a\in A$ with $f(a)=b$. You are given the fact, that $f(g(t))=t$ for all $t\in B$. Aha: Given $b\in B$, we have $a:=g(b)\in A$ with $f(a)=f(g(b))=b$.
Regarding short exact sequences of groups:
Assume $\alpha(x)=\alpha(y)$. then $\alpha(xy^{-1})=1$, i.e. $xy^{-1}\in\ker(1\to A)=\{1\}$, i.e. $xy^{-1}=1$ and finally $x=y$.
Assume $b\in B$. Then $b\in\ker(B\to 1)$, hence $b\in\operatorname{im}(\beta)$, i.e. there exists $g\in G$ with $\beta(g)=b$.
A: Hints:
For (1)a), if $f$ is not injective, then there exist $a\neq b\in A$ such that $f(a)=f(b)$. Hence $gf(a)=gf(b)$. On the other hand, $id_A(a)=a$ and $id_A(b)=b$. Is it possible that $gf=id_A$?
For (1)b), if $f$ is not surjective, then there exists $b\in B$ such that for all $a\in A$, $f(a)\neq b$. Is it possible that $fg=id_B$? (What happens to $fg(b)$? could it be $b$?)
For (2): By the definition of short exact sequence, you have $\ker(\alpha)=\operatorname{Im}(1)=\{1_A\}$, $\operatorname{Im}(\alpha)=\ker(\beta)$ and $\operatorname{Im}(\beta)=\ker(1)=B$.
Can you prove that $\varphi:G\to H$ is injective if and only if $\ker(\varphi)=\{1_G\}$ 
and is surjective if and only if $\operatorname{Im}(\varphi)=H$?
