# $h:X\to X$ endomorphism of module $X$ over $R$ satisfying $h\circ h = h$. Prove X = Im(h)\oplus Ker(h) [duplicate]

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?

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); 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.

• Why should $h\circ h$ be a bijection? – Bernard Apr 21 '17 at 22:58

## 1 Answer

$x=(x-h(x))+h(x)$, $x-h(x)\in Ker(h), h(x)\in Imh$. 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 $Imh\cap Ker h=0$.