Let $M,N,P$ be $R$-modules, prove that $\text{Hom}_R(M \oplus P,L) \cong \text{Hom}_R(M,L)\oplus \text{Hom}_R(P,L)$.

Attempt:Let $\pi_1:M \oplus P\rightarrow M$ and $\pi_2:M \oplus P \rightarrow P$ be the natural projections. If $f \in \text{Hom}_R(M\oplus P,L)$ then $f \circ \pi_1$ and $f \circ \pi_2$ determine elements in $\text{Hom}_R(M,L)$ and $\text{Hom}_R(P,L)$ respectively. Thus $f:\text{Hom}_R(M \oplus P,L) \rightarrow \text{Hom}_R(M,L)\oplus \text{Hom}_R(P,L)$ given by $f \mapsto (f \circ \pi_1) \times (f \circ \pi_2)$ determines a homomorphism. Let $f_1 \in \text{Hom}_R(M,L),f_2 \in \text{Hom}_R(P,L)$, and define $f \in \text{Hom}_R(M \oplus P,L)$ by $f=f_1 \oplus f_2$. Then we have that $\text{Hom}_R(M,L)\oplus \text{Hom}_R(P,L) \rightarrow \text{Hom}_R(M \oplus P,L)$ given by $f_1 \times f_2 \mapsto f_1 \oplus f_2$ is the inverse of the first map. Let $\Phi$ be the first map, and $\Psi$ the second. We have \begin{equation} \begin{split} (\Phi \circ \Psi)(f_1 \times f_2))=&\Phi \circ (f_1 \oplus f_2)\\ =&((f_1 \oplus f_2) \circ \pi_1) \times ((f_1 \oplus f_2) \circ \pi_2))\\ =&f_1 \times f_2 \\ \\ (\Psi \circ \Phi)(f) =& \Psi \circ ((f \circ \pi_1) \times (f \circ \pi_2))\\ =&(f\circ \pi_1) \oplus (f \circ \pi_2)\\ =&f \end{split} \end{equation}

Is this correct? I believe I should be showing the inverses by looking at the images of specific elements in the modules themselves, to do this. What is a better/correct way to solve this problem?


1 Answer 1


The idea is correct, but there are some small issues. For example, how do you define $f_1 \oplus f_2$? Usually when you have $f_1 : M_1 \to N_1$, $f_2 : M_2 \to N_2$, the map $f_1 \oplus f_2 : M_1 \oplus M_2 \to N_1 \oplus N_2$ is defined by applying $f_1$, $f_2$ componentwise. Furthermore the compositions $f \circ \pi_1, f \circ \pi_2$ do not make sense, you want $f \circ j_1, f \circ j_2$, with $j_1, j_2$ the inclusions $M \to M \oplus P, P \to M \oplus P$.

So, instead:

Define $$\Phi: \mathrm{Hom}_R(M \oplus P, L) \to \mathrm{Hom}_R(M, L) \oplus \mathrm{Hom}_R(P, L)$$ by sending $f \in \mathrm{Hom}_R(M \oplus P, L)$ to the pair $(f \circ j_1, f\circ j_2)$.

Conversely, given $f_1 \in \mathrm{Hom}_R(M, L)$, $f_2 \in \mathrm{Hom}_R(P, L)$, define $f : M \oplus P \to L$ by $f(m, p) = f_1(m) + f_2(p)$, giving a homomorphism $$\Psi: \mathrm{Hom}_R(M, L) \oplus \mathrm{Hom}_R(P, L) \to \mathrm{Hom}_R(M \oplus P, L).$$

I'll leave verifying that both of these are homomorphisms and that they are inverses to each other to you.


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