Dimension of cokernel of composition of maps

Question

Let $F:X\to Y$ and $G:Y\to Z$ be two linear maps between vector spaces.

How do you show that $$ \dim(\mathrm{coker}(G\circ F))) \leq \dim(\mathrm{coker}(F))+\dim(\mathrm{coker}(G)) $$?

It is clear that $\mathrm{im}(G\circ F) \subseteq \mathrm{im}(G)$ because $F(X)\subseteq Y$. This gives the intuitive notion that $\mathrm{coker}(G\circ F) \supseteq \mathrm{coker}(G)$ (morally), and so $\dim(\mathrm{coker}(G))\leq\dim(\mathrm{coker}(G\circ F)$. How to proceed from here?

There is this answer: https://math.stackexchange.com/a/1113704/61151 but it doesn't go into enough detail (e.g. how does one get the S.E.S. and how to infer from them a formula on the dimensions).

Answer 1

Given $F \colon X \rightarrow Y$, you have

$$ \dim \mathrm{coker}(F) = \dim Y - \dim \mathrm{im}(F) = \dim Y - (\dim X - \dim \ker(F)) \\ = \dim \ker(F) + \dim Y - \dim X. $$

Thus, we have

$$ \dim \mathrm{coker}(G \circ F) = \dim \ker(G \circ F) + \dim Z - \dim X $$

while

$$ \dim \mathrm{coker}(G) + \dim \mathrm{coker}(F) = \\ \dim \ker(G) + \dim Z - \dim Y + \dim \ker(F) + \dim Y - \dim X = \\ \dim \ker(G) + \dim \ker(F) + \dim Z - \dim Y.$$

Now use the inequality for the dimension of the kernel for composition of linear maps (asked before here).

If you want a method of proof that works for $X,Y,Z$ that are (possibly) infinite dimensional, choose ${y_1, \ldots, y_m}$ such that $Y = \mathrm{im}(F) \oplus \mathrm{span} \{ y_1, \ldots, y_m \}$. Then, $\dim \mathrm{coker}(F) = m$. Similarly, choose ${z_1, \ldots, z_l}$ such that $Z = \mathrm{im}(G) \oplus \mathrm{span} \{ z_1, \ldots, z_l \}$. Then $\dim \mathrm{coker}(G) = l$. Show that $Z = \mathrm{im}(G \circ F) + \mathrm{span} \{ G(y_1), \ldots, G(y_m), z_1, \ldots, z_l \}$. This will imply that $\{ [G(y_1)], \ldots, [G(y_m)], [z_1], \ldots, [z_l] \}$ is a spanning set for $Z / \mathrm{im}(G \circ F)$ showing that $\dim \mathrm{coker}(G \circ F) \leq m + l = \dim \mathrm{coker}(F) + \dim \mathrm{coker}(G)$.