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This question already has an answer here:

Let $h:X\to X$ denote an endomorphism of a module $X$ over $R$ satisfying $h\circ h = h$. Prove

$$X = Im(h)\oplus Ker(h)$$

I have this theorem:

If the composition $h=g\circ f$ of two homomorphisms $f:X\to Y$ and $g:Y\to Z$ of modules $X,Y,Z$ over $R$ is an isomorphism, then the following statements hold:

i) $f$ is a monomorphism

ii) $g$ is an epimorphism

iii) The module $Y$ is decomposable into the direct sum of $Im(f)$ and $Ker(g)$

Wel, the composition $h\circ h$ is surely a bijection, and since i'ts an endomorphism, it's a homomorphism. So should it follow directly from this theorem?

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marked as duplicate by rschwieb abstract-algebra Apr 22 '17 at 2:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Why should $h\circ h $ be a bijection? $\endgroup$ – Bernard Apr 21 '17 at 22:58
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$x=(x-h(x))+h(x)$, $x-h(x)\in Ker(h), h(x)\in Imh$. Suppose $y\in Kerh\cap Im h$, there exists $x$ such that $h(x)=y$ since $y\in Im(h)$, we have $h(y)=0=h^2(x)=h(x)=y$. This implies that $Imh\cap Ker h=0$.

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