Prove the sequence is exact: 0 → ker(f) → V → im(f) → 0

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Problem:
Let f ∶ V → V be a linear operator on a finite-dimensional vector space V .
Prove that the sequence 0 → ker(f) → V → im(f) → 0 is exact at each term.

Attempt:
If I call:
  • a: 0 → ker(f),
  • b: ker(f) → V,
  • c: V → im(f),
  • d: im(f) → 0.
Then the sequence is exact at:
  • ker(f) if im(a)=ker(b),
  • V if im(b)=ker(c),
  • im(f) if im(c)=ker(d).

I can show this for the following:

At im(f):
im(c)={c(v) | v∈V }=im(f)={e∈im(f) | d(e)=0}=ker(d)

At V:
im(b)={b(v) | v∈ker(f)}=ker(f)={e∈V | c(e)=0 ∈ im(f)}=ker(c)

I'm having trouble showing it is exact at ker(f) though. I know that im(a)=0 and ker(b)={v∈ker(f) | b(v)=0 ∈V} however doesn't this mean that ker(b)=ker(f), and so how can I deduce that ker(b)=ker(f)=im(a)=0?

Thanks for any help!
 

Answers and Replies

fresh_42
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You must have a look on the definition of ##b : \ker(f) \rightarrow V##. E.g. if we defined ##b =0## then ##b## would still be linear and a map from ##\ker (f)## to ##V##. So the way how ## b## is defined is crucial here.
 

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