If $M_1$ and $M_2$ are R-modules, $R$ possibly non-commutative. Show that $M_1 \oplus M_2$ is projective $\iff M_1$ and $M_2$ are both projective
I have no idea how to solve this. By definition an $R$-module is defined to be projective if the functor $Hom_R(P,−): RMod \to Ab$ is exact. Any help is appreciated
On
Theorem. $\mathrm{Hom}_R(P,-)\colon \mathsf{RMod}\to\mathsf{Ab}$ is exact if and only if for every surjection $f\colon M\to N$ in $\mathsf{RMod}$ and every $h\colon P\to N$, there exists $g\colon P\to M$ such that $fg=h$.
Proof. Assume $\mathrm{Hom}_R(P,-)$ is exact. Let $f\colon M\to N$ be surjective, and $h\colon P\to N$ be any morphism.
Since $\mathrm{Hom}_R(P,-)$ is exact, $\mathrm{Hom}_R(P,f)\colon \mathrm{Hom}_R(P,M)\to\mathrm{Hom}_R(P,N)$ is surjective. Therefore, there exists $g\in\mathrm{Hom}_R(P,M)$ such that $\mathrm{Hom}_R(P,f)(g)=h$; but $\mathrm{Hom}_R(P,f)(g)=fg$, so there exists $g\colon P\to M$ such that $fg=h$.
Conversely, assume the property holds. To show the functor is exact it suffices to show that if $A\stackrel{f}{\to}B\stackrel{g}{\to}C$ are left $R$-modules and module homomorphisms such that $\mathrm{Im}(f)=\ker(g)$, then $\mathrm{Im}(f\circ -)=\mathrm{ker}(g\circ -)$, where $$f\circ -\colon\mathrm{Hom}_R(P,A)\to\mathrm{Hom}_R(P,B),\qquad g\circ -\colon \mathrm{Hom}_R(P,B)\to\mathrm{Hom}_R(P,C).$$ Since $g\circ f$ is the zero map, $(g\circ -)\circ (f\circ -) = (g\circ f)\circ -$ is the zero map.
Now let $h\colon P\to B$ be such that $g\circ h$ is the zero map. Then $h(P)\subseteq \ker(g)=\mathrm{Im}(f)$. Let $f'\colon A\to\mathrm{Im}(f)$ be the correstriction of $f$, $h'\colon P\to \mathrm{Im}(f)$ be the correstriction of $h$, and let $i\colon\mathrm{Im}(f)\hookrightarrow B$ be the inclusion map. Since $f'$ is surjective, the property says that there exists $u\colon P\to A$ such that $f'\circ u = h'$. Therefore, $f\circ u = i\circ f'\circ u = i\circ h'=h$, so $h$ lies in the image of $f\circ -$. Thus, we have $\mathrm{Im}(f\circ -) = \ker(g\circ -)$, as desired. $\Box$
Corollary. For a left $R$-module $P$, the following are equivalent:
Proof. The equivalence of 1 and 2 is the Theorem. The equivalence of 2, 3, and 4 is proven here.
A module satisfying any (and hence all) of the above four conditions is said to be "projective."
Corollary. Let $M_1$ and $M_2$ be $R$-modules. Then $M_1\oplus M_2$ is projective if and only if $M_1$ and $M_2$ projective.
Proof. If $M_1$ and $M_2$ are projective, then there exist $Q_1$ and $Q_2$ sucht hat $M_i\oplus Q_i$ is free, and then $(M_1\oplus M_2)\oplus (Q_1\oplus Q_2)$ is a direct sum of free modules, hence free. Converseley, if $M_1\oplus M_2$ is projective, there exists $Q$ such that $(M_1\oplus M_2)\oplus Q = F$ is free. Then $M_1$ is free, witnessed by $M_1\oplus (M_2\oplus Q)=F$, and $M_2$ is free, witnessed by $M_2\oplus (M_1\oplus Q)\cong F$. $\Box$
On
I want to include a proof of Daniel's initial answer.
Take $L\in Hom(M_1\oplus M_2,X)$, then define:
$$[\color{purple}{\eta_{X}}(L)]:= (L_1,L_2)$$
where $\color{purple}{L_1}(m_1) := L((m_1,0))$ and $\color{purple}{L_2}(m_2) := L((0,m_2)).$
These are right $R$-linear maps as $L$ is AND operations are component-wise in $M_1\oplus M_2$ by definition.
Since the codomain of $L$ is $X$, each $L_i$ takes image in $X$. So we have well definition:
$$\eta_X(L)\in Hom(M_1,X)\times Hom(M_2,X).$$
To show this map is invertible, define:
$$\color{purple}{\eta_X^{-1}}((L_a,L_b)) := L'$$
where $\color{purple}{L'}((m_1,m_2)):= L_a(m_1)+L_b(m_2)$.
This is a right $R$-linear map due to component wise operations in the direct sum AND because the $L_i$ are such maps. Because the codomain of the $L_i$'s is $X$ and $X$ is closed under its own $+$ operation, the image of $L'$ is in $X$. So $$\eta_X^{-1}((L_a,L_b))\in Hom(M_1\oplus M_2,X).$$
Now, $$[\eta_X^{-1} \circ \eta_X(L)](m_1,m_2) = L_1(m_1)+L_2(m_2) = L((m_1,0)+(0,m_2)) = L((m_1,m_2))$$
and likewise:
$$\eta_X\circ \eta_X^{-1}(L_a,L_b) = (L_1,L_2),$$
where $L_1(m_1):= (L_a+L_b)((m_1,0)) = L_a(m_1)+L_b(0) = L_a(m_1)$ and similarly $L_2(m_2) = L_b(m_2)$.
Hence both compositions yield their respective identities and $$\eta_X:Hom(M_1\oplus M_2,X)\to Hom(M_1,X)\times Hom(M_2,X)$$
is an isomorphism for arbitrary $X$.
It remains to check the following commutativity conditions are satisfied, to make $\eta$ natural: $$\color{blue}{\big[\eta_Y\circ F(f) = G(f)\circ \eta_X\big](Hom(M_1\oplus M_2,X))}$$ with $$\color{purple}{F(X)}:= Hom(M_1\oplus M_2,X)\text{ }\text{ and }\text{ }\color{purple}{G(X)}:= Hom(M_1,X)\times Hom(M_2,X).$$ Accordingly, inserting the argument: $$(\eta_Y\circ F(f))(Hom(M_1\oplus M_2,X)) = \eta_Y(Hom(M_1\oplus M_2,Y))$$ $$= Hom(M_1,Y)\times Hom(M_2,Y)$$ $$= G(f)\big(Hom(M_1,X)\times Hom(M_2,X)\big)$$ $$= (G(f)\circ \eta_X)(Hom(M_1\oplus M_2,X)).$$
We conclude $Hom(M_1\oplus M_2,\cdot)\cong Hom(M_1,\cdot)\times Hom(M_2,\cdot)$ as functors in $[Mod_R,Ab]$. $\square$
As for the other remark. Consider a short exact sequence: $$0\to X \xrightarrow{\color{purple}{\alpha}} Y \xrightarrow{\color{purple}{\beta}} Z\to 0$$ We know exactness at each node implies respectively: $$\color{darkblue}{ker(\alpha) = \{0\},\text{ }Im(\alpha)=ker(\beta),\text{ and }Im(\beta) = cod(\beta)}.$$ Since the direct product of morphisms just direct products the domains and codomains, we have $ker$ and $Im$ separate as desired: $$ker\big(Hom(M_1\oplus M_2,\gamma)\big) = ker\big(Hom(M_1,\gamma)\big)\times ker\big(Hom(M_2,\gamma)\big)$$ $$\text{ and }$$ $$Im\big(Hom(M_1\oplus M_2,\gamma)\big) = Im\big(Hom(M_1,\gamma)\big)\times Im\big(Hom(M_2,\gamma)\big).$$ From this, it is apparent that the three statements above in $\color{darkblue}{\text{darkblue}}$ are satisfied by $Hom(M_1\oplus M_2,\gamma)$ iff they are satisfied by both $Hom(M_i,\gamma)$, using $\gamma := \alpha\text{ or }\beta$.
So we see that exactness transfers from $F$ to and from the constituent functors. Finally proving $M_1\oplus M_2$ projective iff $M_1$ and $M_2$ are projective. $\square$
You use the universal property of the coproduct (i.e. direct sum in our case) to see that $\mathrm{Hom}_R(M_1\oplus M_2,-)\cong\mathrm{Hom}_R(M_1,-)\times\mathrm{Hom}_R(M_2,-)$ as functors $\mathrm{Mod}_R\to\mathrm{Ab}$. Now, a product of functors is exact iff all the functors involved are exact. This is essentially immedate from the proposition in this question, which is not hard to prove by checking things on elements. (And I gave in there also a proof which tries to avoid elementwise arguments as much as possible.)
Edit: for a proof using that a module is free if it has left lifting against surjections, we first prove that if $M_1$ and $M_2$ are both projective, so is their sum $M_1\oplus M_2$. Suppose given a surjection $g\colon N\twoheadrightarrow N'$, and a map $h\colon M_1\oplus M_2\to N'$. Such a map $h$ is of the form $h(m_1,m_2)=h_1(m_1)+h_2(m_2)$ for some $h_i\colon M_i\to N'$. Since $M_1$ and $M_2$ are projective, we can lift either $h_i$ to a map $\widetilde{h}_i\colon M_i\to N$ such that $g\widetilde{h}_i=h_i$. Define $\widetilde{h}\colon M_1\oplus M_2\to N, (m_1,m_2)\mapsto \widetilde{h}_1(m_1)+\widetilde{h}_2(m_2)$. Then this is an $R$-linear map, and $g\widetilde{h}=h$. Therefore, $M_1\oplus M_2$ is projective.
Conversely, if $M_1\oplus M_2$ is projective, we will prove that $M_1$ is projective. For a surjection $g\colon N\twoheadrightarrow N'$ and a map $h\colon M_1\to N'$, we define a map $p_1\colon M_1\oplus M_2\to M_1, (m_1,m_2)\mapsto m_1$ and consider the composition $k:=hp_1\colon M_1\oplus M_2\to N'$. We can lift this to a map $\widetilde{k}\colon M_1\oplus M_2\to N$ such that $g\widetilde{k}=k$, because $M_1\oplus M_2$ is projective. Now put $\widetilde{h}\colon M_1\to N, m_1\mapsto \widetilde{k}(m_1,0)$. Then $g\widetilde{h}(m_1)=g\widetilde{k}(m_1,0)=hp_1(m_1,0)=h(m_1)$, so $\widetilde{h}$ is the desired lift that shows that $M_1$ is projective.