# $\dim(V_1) - \dim(V_2) + \dim(V_3) - \dim(V_4) = 0$

Let $V_1$, $V_2$, $V_3$, $V_4$ be linear spaces over the same field K

$$f_1: V_1 \to V_2 , f_2: V_2 \to V_3 , f_3: V_3 \to V_4 , f_4: V_4 \to V_1$$

with

$$\mbox{im}(f_1) = \ker(f_2) , \mbox{im}(f_2) = \ker(f_3) , \mbox{im}(f_3) = \ker(f_4) , \mbox{im}(f_4) = \ker(f_1)$$

Show that:

$$\dim(V_1) - \dim(V_2) + \dim(V_3) - \dim(V_4) = 0$$

I know I have to show it by applying dimension rules, but I have no idea how to practically do it.

• this is why it is nice to know what an exact sequence is ;) – Alexander Grothendieck Nov 16 '13 at 19:18

$dim(V_{1}) - dim(V_{2}) + dim(V_{3}) - dim(V_{4}) = (im(f_{1})+ker(f_{1}))-(im(f_{2})+ker(f_{2})) + (im(f_{3}) + ker(f_{3})) - (im(f_{4})+ker(f_{4})) = im(f_{1}) + im(f_{4}) - im(f_{2}) - im(f_{1}) + im(f_{3}) + im(f_{2}) - im(f_{4}) - im(f_{3}) = 0$.