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So the question is as follows:

Let $0, \mathbb{{C}}^{n}, \mathbb{{C}}^{\ell}$ and $\mathbb{{C}}^r$ be $\mathbb{{C}}$-vector spaces (where $0$ is the trivial vector space), and let $f_i$, $(i=0,…,3)$, be $\mathbb{{C}}$-linear maps: $$0 \xrightarrow{f_0} \mathbb{{C}}^{n} \xrightarrow{f_1} \mathbb{{C}}^{\ell} \xrightarrow{f_2} \mathbb{{C}}^{r} \xrightarrow{f_3} 0$$ satisfying $\ker f_{i+1} =\operatorname{im} f_i$ for $i=0,1,2$.

Any tips on how to find the value of ${\ell}$?

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Since the sequence is exact, $f_1$ is injective. So $\dim\operatorname{im}f_1= n$. By exactness at $\mathbb{C}^l$, we know that $\dim\ker f_2=n$. By rank-nullity for $f_2$, we know that $\dim\ker f_2+\dim\operatorname{im}f_2=l$. But since $f_2$ is surjective, $\dim\operatorname{im}f_2= r$. So $l=n+r$.

In other words, every exact sequence in the category of finite dimensional $\mathbb{C}$-vector spaces is split exact.

  • Your category has only finite-dimensional spaces in it? Or not? – GEdgar Nov 19 '20 at 00:52
  • I don't think the last line is really relevant here. – mrtaurho Nov 19 '20 at 00:53
  • @GEdgar nLab claims so too (without any restrictions on the dimension). – mrtaurho Nov 19 '20 at 00:56
  • @mrtaurho I'm not sure, but my understanding is that every exact sequence splits even without assuming finite dimensional. If I am wrong, please let me know. –  Nov 19 '20 at 00:56
  • @LAGC You're correct assuming the general case (especially the religious part ;) ). The finite dimensional case follows from rank-nullity, the general using the axiom of choice by picking bases (Relevant). – mrtaurho Nov 19 '20 at 00:58
  • @mrtaurho Ah I see. Choice bothers me not, but indeed as you say the general remark is not relevant to the proof of the problem that I provided. –  Nov 19 '20 at 01:00
  • You say "In other words", so you are claiming that your proof shows this fact. – GEdgar Nov 19 '20 at 01:01
  • @GEdgar Yes, I agree, that is not correct of me to say. I will edit to reflect. –  Nov 19 '20 at 01:02