En primer lugar, si $\text{coker}\ f$ es Artinian, entonces también lo es $\text{coker}\ f^m$ todos los $m\geq 1$: basta para comprobar esto $m=2$, y en este caso, tenga en cuenta que hay un epimorphism $$\text{coker}\ f\twoheadrightarrow\text{im} f / \text{im} f^2$$ and a short exact sequence $$0\to\text{im}\ f / \text{im}\ f^2\to\text{coker}\ f^2\to \text{coker}\ f\to 0.$$It remains to show that all $\texto{ker}\ f^m$ are Artinian. For that, note that since $\text{ker}(f^n)\subseteq\text{ker}(f^m)$ for $m\geq n$ (and, as we have already seen, all $\text{coker}\ f^m$ are Artinian again), we may without loss of generality replace $f$ by $f^m$ for any $m\geq 1$. Now since $X$ is Noetherian, there is $m\gg 0$ such that $\text{ker}\ f^m=\text{ker}\ f^{m+1}$, so replacing $f$ by $f^m$ we may therefore assume right from the beginning that $\text{ker}\ f=\text{ker}\ f^2$. Then $\text{ker}(f)\cap\text{im}(f)=0$, hence $\text{ker}\ f\hookrightarrow X\twoheadrightarrow\text{coker}\ f$ is injective and $\text{ker}\ f$ es Artinian como un subobjeto de un Artinian objeto.
