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Supoose $U$, $V$ and $W$ are vector spaces. Let $f: U\to V$ and $g:V\to W$ are two linear transformations. Do we have any formulas expressing the dimension of $\operatorname{coker}(gf)$ in terms of dimensions of $\operatorname{coker}(f)$ and $\operatorname{coker}(g)$?

messi
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  • Since $\mbox{im},(gf)\subseteq\mbox{im},(g)$, you certainly have that $\dim\mbox{coker},(g)\leq\dim\mbox{coker},(gf)$ – A. Bellmunt Jun 05 '13 at 10:25
  • Consider that there's no formula for $\ker(gf)$ in terms of $\ker(g)$ and $\ker(f)$. – egreg Jun 05 '13 at 10:49

2 Answers2

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This is an old question but since I was thinking about the same question, let me give you a bit more positive answer:

More generally let $U,V,W$ be left modules over a ring $R$ with one and the maps are $R$-linear. Then there is a right exact sequence $$ \operatorname{coker} f \longrightarrow \operatorname{coker}gf \longrightarrow \operatorname{coker} g \longrightarrow 0.$$ Under the additional hypothesis that $g$ is injective, we even get a short exact sequence $$0 \longrightarrow \operatorname{coker} f \longrightarrow \operatorname{coker}gf \longrightarrow \operatorname{coker} g \longrightarrow 0.$$ In particular you get some information about the dimensions in the vector space case. In general $$ \dim \operatorname{coker} g \le \dim\operatorname{coker} gf \le \dim \operatorname{coker} f + \dim\operatorname{coker} g.$$ When $g$ is injective, you get $$\dim\operatorname{coker} gf = \dim \operatorname{coker} f + \dim\operatorname{coker} g.$$

One can even pin the dimension down further with the following information. One should note that $$\ker( \operatorname{coker} f \to \operatorname{coker} gf) \cong (\operatorname{im} f +\ker g)/ \operatorname{im} f\cong \ker g/(\operatorname{im}f\cap \ker g).$$ Thus $$\dim\operatorname{coker} gf = \dim \operatorname{coker} f + \dim\operatorname{coker} g-\dim (\operatorname{im}f+\ker g) + \dim \operatorname{im}f$$ and $$\dim\operatorname{coker} gf = \dim \operatorname{coker} f + \dim\operatorname{coker} g-\dim \ker g + \dim (\operatorname{im}f\cap \ker g).$$

Peter Patzt
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  • Sorry for upping an old post, but which maps are denoted by the arrows in the s.e.s. of cokernels? – Luigi M Mar 30 '15 at 09:48
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    To be a bit more elaborate, $\operatorname{coker} f = V/\operatorname{im} f$, $\operatorname{coker} gf = W/\operatorname{im} gf$, $\operatorname{coker} g = W/\operatorname{im} g$. Since $g$ maps $\operatorname{im} f$ to $\operatorname{im} gf$, $g$ induces a map $\operatorname{coker} f \to \operatorname{coker} gf$. Since $\operatorname{im} gf \subset \operatorname{im} g$, there is a surjection $W/\operatorname{im} gf \to W/\operatorname{im} g$. The kernel of this surjection is $\operatorname{im} g/\operatorname{im} gf$ which is exactly the image of the first map. Hope that helps. – Peter Patzt Mar 30 '15 at 13:42
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There can be no such formula. To see why, let $V = U \cong \mathbb{F}^2$, where $\mathbb{F}$ is the base field, and let $W = \mathbb{F}$. Take a basis $\{e_1,e_2\}$ for $U$, and let $g : e_1 \mapsto 1, e_2 \mapsto 0$. Now consider two different choices for $f$:

  1. $f_1 : e_1 \mapsto e_1, e_2 \mapsto 0$.
  2. $f_2 : e_1 \mapsto e_2, e_2 \mapsto 0$.

In each case, the cokernel of $f_i$ has dimension one, but the cokernel of $gf_1$ is $0$, whereas the cokernel of $gf_2$ is $\mathbb{F}$.

Rhys
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