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?


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.

$\hskip1.9in$enter image description here

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*}

  • $\begingroup$ Diagram, you say? ;) (not required) $\endgroup$ – BrianO Oct 30 '15 at 18:44
  • $\begingroup$ Working on it! :) $\endgroup$ – Viktor Vaughn Oct 30 '15 at 18:45
  • 1
    $\begingroup$ Good show! (and good answer too) $\endgroup$ – BrianO Oct 30 '15 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.