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July 7th, 2014, 12:11 PM  #1 
Newbie Joined: Jul 2014 From: Europe Posts: 1 Thanks: 0  Direct sum of HomSets
I am struggling to prove the following statement: Hom(V,W_1 + W_2) = Hom(V,W_1) + Hom(V,W_2), where + denotes the direct sum, V, W_1 and W_2 are Gmodules and V is irreducible. Hom(X,Y) denotes the set of homomorphisms from X to Y. I have already looked elsewhere, but I just found explanations involving category theory that I do not understand. Is there an easy explanation, why the statement is true? Thank you very much for your help! 
July 7th, 2014, 02:29 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
I've never seen this before, but I bet I can figure it out. Suppose we have the maps: $\pi_1:W_1\oplus W_2 \to W_1$ $\pi_2:W_1\oplus W_2 \to W_2$ given by: $\pi_1(w_1,w_2) = w_1$ (some texts write $(w_1,w_2)$ as $w_1+w_2$ or $w_1\oplus w_2$) $\pi_2(w_1,w_2) = w_2$. Let's verify that these are in $\text{Hom}(W_1\oplus W_2,W_1)$ and $\text{Hom}(W_1\oplus W_2,W_2)$, respectively: $\pi_1((w_1,w_2) + (w_1',w_2')) = \pi_1((w_1+w_1',w_2+w_2')) = w_1 + w_1' = \pi_1((w_1,w_2)) + \pi_1((w_1',w_2'))$ so we have an abelian group homomorphism. $\pi_1(g\cdot (w_1,w_2)) = \pi_1((g\cdot w_1,g\cdot w_2)) = g \cdot w_1 = g\cdot \pi_1(w_1,w_2)$, so $\pi_1$ is equivariant. The proof for $\pi_2$ is just the same, I'm sure you can devise your own. Now what we need to come up with is a $G$module isomorphism: $f:\text{Hom}(V,W_1\oplus W_2) \to \text{Hom}(V,W_1)\oplus \text{Hom}(V,W_2)$. So, given $\eta \in \text{Hom}(V,W_1\oplus W_2)$, define: $f(\eta) = (\pi_1\circ\eta,\pi_2\circ\eta) \stackrel{\text{def}}{\equiv}(\pi_1\circ\eta) + (\pi_2\circ\eta)$ We need to show 2 things: 1. $f$ is a $G$homomorphism 2. $f$ is a bijection This should get you started, if you have trouble I'll post more later. I believe you'll wind up using the irreducibility of $V$ in showing $f$ is bijective. 

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direct, homsets, sum 
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