If $0 \stackrel{\gamma}\to C_n \stackrel{d_n}\to C_{n-1} \to \dots C_0\to 0$ is exact, then it induces the short exact sequence
$$0 \to \ker d_n \to C_n \to \text{img}d_{n} \to 0?$$
This is part of the Euler Characteristic proof. I stared at this for a while, but it isn't too clear to me. Is this the correct justification?
Since $\ker d_n \hookrightarrow{} C_n$ in a one-to-one fashion, the boundary map $C_n \stackrel{d_{n}}\to\text{img}d_{n} \subset C_{n-1}$ maps back into $C_{n-1}$ (actually it is $0$ from $d^2 = 0$?). So $\ker d_n \subset \text{img}d_{n}$. Also $\text{img}d_{n} \approx C_n/\ker d_n = C_n/\text{img}\gamma_{n} = C_n/\{0\} \approx C_n$. Actually I am not sure how to find the other inclusion.
If $A\stackrel f\to B$ is a homomorphism, then $$0\to \ker f\stackrel \subseteq\to A \stackrel f\to\operatorname{img}f \to 0$$ is a short exact sequence. Clearly, the incusion is injective and $f$ is surjective (to $\operatorname{img}f$). Finally the kernel of $f$ is ... $\ker f$, as desired.