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$\DeclareMathOperator{\Hom}{Hom}$ I am working on the problem below.

Let $A,B$ and $M$ be $R-$modules. Show that

(1) $\Hom_R(A\times B,M)\cong \Hom_R(A,M)\times \Hom_R(B,M)$.

For (1), I built a homomorphism $F:\Hom_R(A,M)\times \Hom_R(B,M) \rightarrow \Hom_R(A \times B,M)$ defined by $F(\varphi_1,\varphi_2)=\varphi_1+\varphi_2$.

It is well defined since $\varphi_1+\varphi_2=\psi_1+\psi_2$ whenever $(\varphi_1,\varphi_2)=(\psi_1,\psi_2)$.

Also, it is homomorphism since, $\forall r\in R$ $\forall (\varphi_1,\varphi_2),(\psi_1,\psi_2)\in \Hom_R(A,M)\times \Hom_R(B,M)$,

\begin{align*} F((\varphi_1(a),\varphi_2(b))+r(\psi_1(a),\psi_2(b)))&=(\varphi_1(a)+r\psi_1(a))+(\varphi_2(b)+r\psi_2(b))\\ &=(\varphi_1(a)+\varphi_2(b))+r(\psi_1(a)+\psi_2(b))\\ &=F(\varphi_1(a),\varphi_2(b))+rF(\psi_1(a),\psi_2(b)). \end{align*}

$\forall (a,b)\in A\times B$.

Let $\Phi\in \Hom_R(A\times B,M) $ be given and note that $\Phi(\cdot,0)\in \Hom_R(A,M)$ and $\Phi(0,\cdot)\in \Hom_R(B,M)$, and that for any $(a,b)\in A\times B$,

\begin{align*} F(\Phi(a,0),\Phi(0,b))=\Phi(a,0)+\Phi(0,b)=\Phi(a,b). \end{align*}

Thus, $F$ is surjective.

Therefore, I only need to show that it is an injection. But I am having trouble in there. I just want to show that $\ker(F)=0$ but it seems there are so many $\varphi\in \Hom_R(A,M)$ and $\psi\in \Hom_R(B,M)$ such that $\varphi+\psi=0$. Should I change the homomorphism I have built? It seems this $F$ is only reasonable one...

I thank for any help in advance.

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    $\begingroup$ You are making an assumption about what homomorphisms look like from $A\times B$. Use your surjectivity argument to show that $A\times B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $A\to M$ and $B\to M$ there will be a unique homomorphism from the coproduct to $M$. $\endgroup$ – John Douma Nov 21 '18 at 17:27
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The map you are defining doesn't make sense. You don't need to worry about well-defined-ness of the map as there aren't any equivalence relations around to muck things up. Whatever you define that map to be at an element, will be fine because it not like there are a whole bunch of representatives of that element that could change the expression depending on which ones you pick.

More to the point however, it doesn't make sense to form the sum $\varphi_1 + \varphi_2$ when the domain of $\varphi_1$ is $A$ and the domain of $\varphi_2$ is $B$. Remember that addition of functions is typically defined pointwise, which can only make sense if those functions share the same domain. If you want to combine those two functions in a different manner than pointwise addition, then you need to indicate that by not using the addition symbol.

I'm not going to spend a lot of time reviewing that argument because it's a bit confusing and hard to read. But I can help guide you through the right process to show the necessary bijection. Given two functions $\phi_1: A \to M$ and $\phi_2: B \to M$, define $f(\phi_1,\phi_2)$ to be the function from $A \times B$ to $M$ defined by $f(\phi_1,\phi_2)(a,b) = \phi_1(a)+\phi_2(b)$. Note that this is not a pointwise sum, and there is no need to argue about well-definedness.

Now we just need to exhibit an inverse map for $f$. In this case, this will be slicker than trying to argue for injectivity and surjectivity of $f$ directly. Given a function $\phi: A \times B \to M$, we define $\phi_1(a) = \phi(a,0)$ and $\phi_2(b) = \phi(0,b)$ for all $a \in A$, $b \in B$. Then we define $g(\phi)$ to be the ordered pair $(\phi_1,\phi_2) \in \operatorname{Hom}_R(A,M) \times \operatorname{Hom}_R(B,M)$. Then we have that $g(f(\phi_1,\phi_2) = g(\phi) = (\phi_1,\phi_2)$ and $f(g(\phi)) = f(\phi_1,\phi_2) = \phi$ and so these maps really are inverses of each other. Therefore we are done.

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Showing injectivity amounts to showing that the inverse map is well defined.

Let $F^{-1}$ be the inverse with $F(\phi) = (\phi_1, \phi_2)$ with $\phi_1(x)=\phi(x, 0)$ and $\phi_2(y) = \phi(0,y)$. This map is obviousy well-defined, so $F$ is injective.

Edit: I realize that assuming $F^{-1}$ exists is begging the question. However, the proposition is easier to prove, in my opinion, starting with what I called $F^{-1}$.

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Define the inclusion maps $i_A:A\to A\times B$ by $i_A(a)=(a,0)$ and $i_B:B\to M$ by $i_B(b)=(0,b)$.

As you mentioned, these are module homomorphisms.

Given $f\in Hom(A\times B,M)$, Let $F(f)=(f\circ i_A, f\circ i_B)$. Since both components are the composition of homomorphisms, each component is a homomorphism.

Given $(\phi, \psi)\in Hom(A,M)\times Hom(B,M)$, define $G(\phi,\psi)$ by $G(\phi,\psi)(a,b)=\phi(a)+\psi(b)$.

Then $(G\circ F)(f)=G(f\circ i_A, f\circ i_B)=f\circ i_A + f\circ i_B$.

$(f\circ i_A + f\circ i_B)(a,b)=f\circ i_A(a)+f\circ i_B(b)=f(a,0)+f(0,b)=f(a,b)$.

Therefore, $(G\circ F)(f)=f$.

$(F\circ G)(\phi,\psi)=F(\phi+\psi)=((\phi+\psi)i_A, ((\phi+\psi)i_B)$.

$(\phi+\psi)i_A(a)=(\phi+\psi)(a,0)=\phi(a)$.

Likewise,$(\phi+\psi)i_B(b)=(\phi+\psi)(0,b)=\psi(b)$.

Therefor, $(F\circ G)(\phi,\psi)=(\phi, \psi)$.

Therefore, $F$ and $G$ are inverses of each other so they are isomorphisms.

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