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

[Apothem]

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!

 

[fresh_42]

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.