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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.

$\\$ 1. Let $f:A\to B$ be a map.


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


  1. $A$, $B$, $G$ 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.

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Hint: For 1. Assume that $f$ was not injective, resp. surjective. For 2. Use the definition of short exact sequence. – Julian Kuelshammer Dec 2 '12 at 9:54
Please rewrite your question. What else do you want to show in the end. It is not clear. – Reader Dec 2 '12 at 10:22
Ick, there's no need to invoke proof by contradiction there. Suppose $g \circ f = \textrm{id}_A$ and $f(a) = f(b)$; we want to show $a = b$. How might we do this with the help of $g$? etc. – Zhen Lin Dec 2 '12 at 10:22
@gendizer You seem to be very unsatisfied with the answers you got on this site. A proper way to thank people here is to upvote their answers and accept one of them as the best answer for you. Try to do this if you want to get more answers. – user26857 Dec 3 '12 at 10:11
Btw, there is no splitting here, so I edited the title according to the question. – user26857 Dec 3 '12 at 10:14
up vote 3 down vote accepted

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 $\beta(g)=b$.

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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 $\operatorname{Im}(\varphi)=H$?

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Thx for the tips, but I know all these rules about image and kernel and $f(a)=f(b)$. My issue is using those things to prove the above problems. – grendizer Dec 2 '12 at 10:40
ok so for 1a. If I assume $f$ is injective then I have $f(a)=f(b)$ and then $gf(a)=gf(b)$. I have that $gf(a)=id_{A}(a)=a$ and $gf(b)=id_{A}(b)=b$. Then $a=b$ and $f$ is injective. 1b. Hm, assuming $f$ surjective I have that $f(a)=b$ and in that case $fg(b)=f(a)=b$. – grendizer Dec 2 '12 at 12:10
im not sure what u mean with how to define injective and surjective. its not $f(a)=f(b) \implies a=b$ for injective, and for surjective its that there exists $b\in B$ such that for all $a\in A$, $f(a)=b$. Not sure if thats it, and if its the case I still dont know how to move. Been trying to use the $ker$ and $im$ to no avail. – grendizer Dec 2 '12 at 13:29
Is it clearer now? – Dennis Gulko Dec 2 '12 at 13:34
yes i think so. Im working on it now. Does it mean my "solution" for 1a and 1b is ok? – grendizer Dec 2 '12 at 15:16

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