Proof that the Euler characteristic is additive

Solution 1:

Hint. Decompose the exact sequence you start with with many short exact sequences, each one corresponding to the kernel and image of the maps in the first one.

For example, if $f:A\to B$ is a map, you can construct short exact sequences $$0\to\ker f\to A\to\operatorname{im}f\to0$$.

Later. The point of the hint was for you to use the observation provided by your notes themselves. $\def\rk{\operatorname{rk}}\def\im{\operatorname{im}}$Suppose that we have an exact complex $$0\xrightarrow{f_{-1}} X_0\xrightarrow{f_0} X_1\xrightarrow{f_1} X_2\xrightarrow{f_2}\cdots\xrightarrow{f_{n-1}} X_n\xrightarrow{f_n} 0$$ For each $i\in\{-1,\dots,n\}$ we have the map $f_i:X_i\to X_{i+1}$ (letting $X_{-1}=X_{n+1}=0$ for convenience) so we have a short exact sequence $$0\to\ker f_i\to X_i\to\im f_i\to 0$$ so, assuming we know that the rank is additive in short exact sequences, we have $$\rk X_i=\rk\ker f_i+\rk\im f_i.$$ Multiplying this by $(-1)^i$ and summing over $i$ we see then that $$\sum_{i=-1}^n(-1)^i\rk X_i=\sum_{i=-1}^n(-1)^i\rk\ker f_i+\sum_{i=-1}^n(-1)^i\rk\im f_i. \tag{1}$$ Now, the original exact sequence being exact, we have $\ker f_{i+1}\cong\im f_i$ for all $i$, so of course $\rk\ker f_{i+1}=\rk\im f_i$ for all $i$. Using this in the right hand side of (1) we easily see that in fact $$\sum_{i=-1}^n(-1)^i\rk X_i=0.$$ In your case, my $X_i$s are $A$s, $B$s and $C$s, so you just need to rename stuff