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I am working on the following task: Let $V,W$ be $K$-vector space, $a:V\to W$ is a homomorphism and $U \subset V$ a subspace, $U \subset \ker (a)$. Let $b: V/U\to W$, $v+U \mapsto a(v)$ be a homomorphism. Show: $\ker(b) \cong \ker(a)/U$. My idea: $pi: V \to V/U$, $v \mapsto v+U$.

Thanks for your help!

hannah2002
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1 Answers1

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Hint : Show that the homomorphism $$\pi_{|\mathrm{Ker}(a)} : \mathrm{Ker}(a) \rightarrow \mathrm{Ker}(b)$$

is well-defined, surjective, and that its kernel is $U$. It will then induce an isomorphism $$\tilde{\pi}_{|\mathrm{Ker}(a)} : \mathrm{Ker}(a)/U \rightarrow \mathrm{Ker}(b)$$

TheSilverDoe
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  • Thanks vor your help! But I don´t understand what the mapping rule is: So i take an element of Ker(a) and in which way is it connected to ker(b)? – hannah2002 Feb 04 '21 at 10:33
  • @hannah2002 What I denote $\pi_{|\mathrm{Ker}(a)}$ is just the restriction of the projection map $\pi : V \rightarrow V/U$ to $\mathrm{Ker}(a)$. I am saying that the image of this restriction is equal to $\mathrm{Ker}(b)$ (and I let you prove it !). Then what I use is a very general result stating that if $f : E \rightarrow F$ is an homomorphism, then $E/\mathrm{Ker}(f)$ is isomorphic to $\mathrm{Im}(f)$. – TheSilverDoe Feb 04 '21 at 10:37
  • Now I understand it, thanks. But how can I show that the restriction of the projection map is well-defined? It is enough to check if it is linear and representative independent? – hannah2002 Feb 04 '21 at 10:44
  • @hannah2002 Yes, and also that its image is rightly included in $\mathrm{Ker}(b)$ :) – TheSilverDoe Feb 04 '21 at 10:45
  • Thank you very much! Should I show that Ker(a) is not empty? – hannah2002 Feb 04 '21 at 10:52
  • @hannah2002 It cannot be empty ! As a subspace, it always contains $0$. – TheSilverDoe Feb 04 '21 at 10:53
  • do you have got a hint for me to show that the kernel is U.. thank you! – hannah2002 Feb 04 '21 at 11:46
  • @hannah2002 Just write the definition. $x \in \mathrm{Ker}(\pi_{|\mathrm{Ker(a)}})$ means that $\pi_{|\mathrm{Ker(a)}}(x)=0$. So it means, because $\pi_{|\mathrm{Ker(a)}}$ is the restriction of $\pi$, that $x \in \mathrm{Ker}(\pi)$. Now, what is $\mathrm{Ker}(\pi)$ ? – TheSilverDoe Feb 04 '21 at 12:03