For $0 \xrightarrow{f_0} \mathbb{C}^n \xrightarrow{f_1} \mathbb{C}^\ell \xrightarrow{f_2} \mathbb{C}^r \xrightarrow{f_3} 0$, what is $\ell$?

Question

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 ${\ell}$?

Answer 1

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.