Exercise 9 Chapter IV Section 1 - T W Hungerford, Algebra

Solution 1:

To complete your solution using exact sequences, note that you have shown that there is a homomorphism $h: Im(f) \to A$ (namely $h = f|_{Im(f)}$) such that $f h = 1|_{Im(f)}$. So, the exact sequence is a split. So, $A = Ker(f) \oplus Im(f)$.

There are, however, simpler solutions. Here is one:

Since $ff =f$, for any $y \in Ker(f) \cap Im(f)$, there is $x \in A$ such that $f(x)=y$ and, since $y \in Ker(f)$, we have $f(y)=0$. Putting all together, we have $$y = f(x)= f(f(x)) = f(y)= 0$$ So $Ker(f) \cap Im(f)=\{0\}$.

Given any $x \in A$, note that $x- f(x) \in Ker(f)$. In fact, $$f(x- f(x)) = f(x) - f(f(x)) = 0$$ It is clear that $f(x) \in Im(f)$ and that $x = (x- f(x))+f(x)$. So we have that $A = Ker(f) \oplus Im(f)$.