I am seeing this kind of question a lot in my class, but each time fail to understand how to think about the question. Why is there an isomorphism between sets of homomorphisms and what is it explicitly?
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$\begingroup$ This follows from the universal property of the coproduct. $\endgroup$ – Viktor Vaughn Oct 30 '15 at 18:21
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$\begingroup$ @SpamIAm, what would the isomorphism be in this case? $\endgroup$ – grayQuant Oct 30 '15 at 18:23
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Let $R$ be a ring and $U_1, U_2, V$ be $R$-modules. First, recall that we have canonical injections into the coproduct: \begin{align*} \iota_1: U_1 &\hookrightarrow U_1 \oplus U_2\\ u_1 &\mapsto (u_1, 0)\\ \iota_2: U_2 &\hookrightarrow U_1 \oplus U_2\\ u_2 &\mapsto (0, u_2) \, . \end{align*}
Given $R$-linear maps $\varphi_1: U_1 \to V$ and $\varphi_2: U_2 \to V$, the universal property of the coproduct states that there is a unique $R$-linear map $\varphi = \varphi_1 \oplus \varphi_2: U_1 \oplus U_2 \to V$ such that the following diagram commutes.
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Commutativity of the diagram implies that $(\varphi_1 \oplus \varphi_2)(u_1, u_2) = \varphi_1(u_1) + \varphi_2(u_2)$.
Conversely, given an $R$-linear map $\varphi: U_1 \oplus U_2 \to V$, pre-composition by the injections yields maps $U_1 \to V$ and $U_2 \to V$. Explicitly, let $\varphi_1 = \varphi \circ \iota_1 : U_1 \to V$ and $\varphi_2 = \varphi \circ \iota_2 : U_2 \to V$.
Thus the isomorphism you seek is \begin{align*} \hom_R(U_1, V) \oplus \hom_R(U_2,V) &\longrightarrow \hom_R(U_1 \oplus U_2, V)\\ (\varphi_1, \varphi_2) &\longmapsto \varphi_1 \oplus \varphi_2\\ \end{align*} with inverse \begin{align*} \hom_R(U_1 \oplus U_2, V) &\longrightarrow \hom_R(U_1, V) \oplus \hom_R(U_2,V)\\ \varphi &\longmapsto (\varphi \circ \iota_1, \varphi \circ \iota_2) \, . \end{align*}
answered Oct 30 '15 at 18:38
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