I am currently self-studying category theory and I was trying to solve the following problem
Given a filtered set $I$ and a ring $A$, if $(L_i, f_{ji})$ and $(M_i, g_{ij})$ are inductive systems of respectively right and left $A$-modules show that $\varinjlim (L_i \otimes M_i)_{i\in I} \cong \varinjlim L_i \otimes \varinjlim M_i$.
I would like to know if my proof is correct. I am sorry if it's a bit messy, I did my best to condense it.
We aim to build two morphisms $\phi$ and $\psi$ that will be inverse of each others.
Let for $i\in I$ $f_i:L_i \rightarrow \varinjlim L_i$ and $g_i: M_i \rightarrow \varinjlim M_i$ be the canonical morphisms.
Functoriality of tensor product gives that $(L_i \otimes M_i, f_{ji}\otimes g_{ji})$ is an inductive system and we note $r_i: L_i \otimes M_i \rightarrow \varinjlim (L_i \otimes M_i)$ the canonical morphisms.
For $i\in I$, we have $f_i \otimes g_i: L_i\otimes M_i\rightarrow \varinjlim L_i \otimes \varinjlim M_i$ and
\begin{gather}
(f_j\otimes f_i)\circ(f_{ji}\otimes g_{ji})=(f_j\circ f_{ji})\otimes(g_j\circ g_{ji})=f_i\otimes g_i
\end{gather}
Hence it's compatible with the order so there is a unique morphism $\phi: \varinjlim (L_i \otimes M_i) \rightarrow \varinjlim L_i \otimes \varinjlim M_i$ s.t.
\begin{gather}
\phi \circ r_i = f_i \otimes g_i
\end{gather}
Now we define a map $\psi: \varinjlim L_i \otimes \varinjlim M_i \rightarrow \varinjlim (L_i\otimes M_i)$ in the following way,
Since $\varinjlim L_i = \bigg\{ f_i(x_i), x_i\in L_i\bigg\}$ and $\varinjlim M_i = \bigg\{ g_i(x_i), x_i\in M_i\bigg\}$ for any $i,j\in I$ there is a $k\geq i,j$ since $I$ is filtered and we define $\psi:f_i(l_i)\otimes g_j(m_j) \longmapsto r_k(f_{ki}(l_i) \otimes g_{kj}(m_j))$.
We must show that $\psi$ is a morphism and doesn't depend of the choice of $k$.
By linearity of $r_i$ for all $i\in I$ and by extending the definition of $\psi$ to finite sum s.t. for $S$ finite
\begin{gather}
\psi\big(\sum_{s\in S}f_{i_s}(l_{i_s})\otimes g_{j_s}(m_{j_s})\big)=r_k \big(\sum_{s\in S}f_{ki_{s}}(l_{i_s})\otimes g_{kj_{s}}(m_{j_s})\big)
\end{gather}
with $k = \max\big(\big\{i_s, s\in S \big\}\cup\big\{ j_s, s\in S\big\}\big)$ we see that $\psi$ is a morphism.
Recall that for $i \in I$ by construction of the colimit $r_i = p\circ q_i$ where $q_i$ is the injection into the coproduct and $p$ the quotient map $\coprod \rightarrow \coprod \diagup R$. And $R$ is generated by the elements $(q_i(l_i\otimes m_i) – q_j(f_{ji}(l_i)\otimes g_{ji}(m_j))_{j\geq i}$.
Now let $i,j\in I$ and $k,\alpha\geq i,j$.
There exist a $\beta \geq k,\alpha$ hence,
\begin{aligned}
r_k(f_{ki}(l_i) \otimes g_{kj}(m_j)) &= q_k(f_{ki}(l_i) \otimes g_{kj}(m_j)) + R\\
&= q_\beta\circ (f_{\beta k}\otimes g_{\beta k})(f_{ki}(l_i) \otimes g_{kj}(m_j)) + R\\
&= q_\beta(f_{\beta i}(l_i) \otimes g_{\beta j}(m_j)) + R\\
&=q_\beta \circ (f_{\beta\alpha} \otimes g_{\beta\alpha})(f_{ki}(\alpha_i) \otimes g_{kj}(\alpha_j)) + R\\
&= q_\alpha(f_{\alpha i}(l_i) \otimes g_{\alpha j}(m_j)) + R
\end{aligned}
Hence $\psi$ is well defined.
We see that $\psi$ and $\phi$ are inverses since for $i,j\in I$,
\begin{aligned}
\phi\circ\psi(f_i(l_i)\otimes g_j(m_j)) &=_{k\geq i,j} \phi\circ r_k (f_{ki}(l_i) \otimes g_{kj}(m_j))\\
&= (f_k\otimes g_k)(f_{ki}(l_i) \otimes g_{kj}(m_j))\\
&= f_i(l_i)\otimes g_j(m_j)
\end{aligned}
For $i\in I$
\begin{aligned}
\psi\circ\phi(r_i(l_i\otimes m_i))&=\psi\circ (f_i\circ g_i)(l_i\otimes m_i)\\
&= \psi(f_i(l_i)\otimes g_i(m_i))\\
&= r_i(f_{ii}(l_i)\otimes g_{ii}(m_i))\\
&= r_i(l_i\otimes g_i)
\end{aligned}
Therefore,
\begin{gather}
\varinjlim (L_i \otimes M_i) \cong \varinjlim L_i \otimes \varinjlim M_i
\end{gather}
Best Answer
$\newcommand{\rmod}{\mathsf{AMod}}\newcommand{\modr}{\mathsf{ModA}}\newcommand{\ab}{\mathsf{Ab}}\newcommand{\C}{\mathsf{C}}$Fix a unital ring $A$. Generalising a bit, we are given small categories $I,J$ and diagrams $L:I\to\rmod$, $M:J\to\modr$ (left and right $A$-modules) and you want to verify there is a (canonical) isomorphism of Abelian groups: $$\varinjlim_IL\otimes\varinjlim_JM\cong\varinjlim_{I\times J}L\otimes M$$
There is an abstract-nonsense way to see this and a slightly more concrete way to see this.
NOTE: Throughout this post $\varinjlim_IL$ - as computed in $\rmod$ - is conflated with $\varinjlim_IL$ - as computed in $\ab$. This is justified because the underlying Abelian group functor $\rmod\to\ab$ is cocontinuous (it has a right adjoint, $G\mapsto\ab(A,G)$ where $a\cdot\varphi:=(x\mapsto\varphi(x\cdot a))$ gives the left $A$-module structure).
The abstract-nonsense way, utilising the Fubini theorem and the fact tensors with one object fixed are cocontinuous (they have a right adjoint):
$$\begin{align}\tag{1}\varinjlim_{I\times J}L\otimes M&\cong\varinjlim_{i\in I}\varinjlim_J(L(i)\otimes M)\\\tag{2}&\cong\varinjlim_{i\in I}(L(i)\otimes\varinjlim_JM)\\\tag{3}&\cong\varinjlim_IL\otimes\varinjlim_JM\end{align}$$
Step $(1)$ is the Fubini theorem. Step $(2)$ uses cocontinuity of $L(i)\otimes(-):\modr\to\ab$. Step $(3)$ uses cocontinuity of $(-)\otimes\varinjlim_JM:\rmod\to\ab$.
With thanks to Lukas Heger for pointing out some new category theory to me, we can recover your original problem from the above. In your case, $J=I$ are the same and $I$ is a filtered category. In which case, the canonical $\Delta:I\hookrightarrow I\times I$ is a (co)final functor. That means the canonical comparison map: $$\varinjlim_I(L\otimes M)\Delta\to\varinjlim_{I\times I}L\otimes M$$Is an isomorphism. From this we recover: $$\varinjlim_{i\in I}L(i)\otimes M(i)\cong\varinjlim_{i\in I}L(i)\otimes\varinjlim_{i\in I}M(i)$$
The reason this functor is final is that for any $(x,y)\in I\times I$, the coslice category $(x,y)/\Delta$ is connected:
This implies the colimit-preservation property by an argument given below.
We can see this more concretely with a clearer isomorphism map.
Let $\lambda_i:L(i)\to\varinjlim_IL$ and $\mu_j:M(j)\to\varinjlim_JM$ denote the legs of the colimit cocone for $i,j\in I,J$. Let $\omega_{i,j}:L(i)\otimes M(j)\to\varinjlim_{I\times J}L\otimes M$ denote the legs of the colimit cocone for $(i,j)\in I\times J$. For $X\in\rmod$ and $Y\in\modr$ let $\tau_{X,Y}:X\times Y\to X\otimes Y$, or just $\tau$, denote the tensor map (of Abelian groups).
Fix a pair $(i,j)\in I\times J$. The maps of Abelian groups: $$L(i)\times M(j)\overset{\lambda_i\times\mu_j}{\longrightarrow}\varinjlim_IL\times\varinjlim_JM\overset{\tau}{\longrightarrow}\varinjlim_IL\otimes\varinjlim_JM$$Are bilinear, and so induce unique maps $\gamma_{i,j}:L(i)\otimes M(j)\to\varinjlim_IL\otimes\varinjlim_JM$.
I claim that the $\gamma_\bullet$ assemble to a cocone under the diagram $L\otimes M$; we accordingly get a unique $\gamma:\varinjlim_{I\times J}L\otimes M\to\varinjlim_IL\otimes\varinjlim_JM$.
$\gamma$ is the ‘canonical comparison map' - we want to check it is an isomorphism.
There is a canonical isomorphism of Abelian groups (since $\times\simeq\oplus$ is a colimit for binary products): $$\pi:\varinjlim_IL\times\varinjlim_JM\cong\varinjlim_{I\times J}L\times M$$Whose inverse is induced by the components $\lambda_\bullet\times\mu_\bullet:L\times M\to\varinjlim_IL\times\varinjlim_JM$.
I now define (the map of sets): $$\delta’= \varinjlim_I L\times\varinjlim_JM\overset{\pi}{\cong}\varinjlim_{I\times J}L\times M\overset{\varinjlim_{I\times J}\tau_{L,M}}{\longrightarrow}\varinjlim_{I\times J}L\otimes M$$ Since $\delta’$ is bilinear, out pops a unique: $$\delta:\varinjlim_IL\otimes\varinjlim_JM\to\varinjlim_{I\times J}L\otimes M$$
Consider: $$\begin{align}\delta\gamma\omega_{i,j}\tau_{L(i),M(j)}&=\delta\tau_{\varinjlim_IL,\varinjlim_JM}(\lambda_i\times\mu_j)\\&=\delta’(\lambda_i\times\mu_j)\\&=\omega_{i,j}\tau_{L(i),M(j)}\end{align}$$For all $i,j$; it follows that $\delta\gamma$ is the identity. Similarly $\gamma\delta$ is found to be the identity, so $\gamma$ is an isomorphism.
Now return to your situation where $I=J$ is filtered. Then for every $i\in I$, the legs $\lambda_i\times\mu_i:L(i)\otimes M(i)\to\varinjlim_{I\in I}L\otimes M$ assemble to a cocone under the diagram $(L\otimes M)\Delta:I\to\ab$ giving a canonical map: $$\kappa:\varinjlim_I(L\otimes M)\Delta\to\varinjlim_{I\times I}L\otimes M$$Which in this case is also an isomorphism because $I$ is filtered.
The (hopefully concrete) composite: $$\varinjlim_{i\in I}L(i)\otimes M(i)\overset{\kappa}{\longrightarrow}\varinjlim_{(i,j)\in I\times I}L(i)\otimes M(j)\overset{\gamma}{\longrightarrow}\varinjlim_{i\in I}L(i)\otimes\varinjlim_{i\in I}M(i)$$Is then the desired isomorphism.
Let's elaborate on why $\kappa$ is an isomorphism:
Let $F:I\to J$ be a functor between the (small) categories $I,J$ with the property that for all $j\in J$, the coslice $j/F$ is connected. Take any category $\C$ and $G:J\to\C$ for which both $\varinjlim_J G$ and $\varinjlim_I GF$ exist.
Let $\lambda_\bullet:GF(\bullet)\to\varinjlim_IGF$ and $\mu_\bullet:G(\bullet)\to\varinjlim_JG$ denote the legs of the colimit cocone. The arrows $(\mu_{F(i)}:GF(i)\to\varinjlim_JG)_{i\in I}$ obviously form a cocone under $GF$ and the "comparison" map $\kappa:\varinjlim_IGF\to\varinjlim_JG$ is so induced.
For all $j\in J$, $j/F$ is nonempty; fix a choice of objects, one for each $j$, i.e. a choice of indices $\iota(j)$ and distinguished arrows $\phi_j:j\to F(\iota(j))$ in $J$.
Define $\varsigma_j:=\lambda_{\iota(j)}\circ G(\phi_j)$ for all $j\in J$. I claim these form a cocone under $G$ (with nadir $\varinjlim_I GF$).
Suppose $f:j\to j'$ is an arrow in $J$. We want to show that $\varsigma_j=\varsigma_{j'}\circ G(f)$, or equivalently that: $$\lambda_{\iota(j)}\circ G(\phi_j)=\lambda_{\iota(j')}\circ G(\phi_{j'}\circ f)$$
I know $\phi_{j'}\circ f$ runs $j\to F(\iota(j'))$: by the connectivity hypothesis, there is a finite zigzag of arrows in $j/F$ connecting $\phi_j$ to $\phi_{j'}\circ f$. So, it remains to show (by a finite induction) that if $\alpha:j\to F(a)$ and $\beta:j\to F(b)$ are directly connected (in a single step) then $\lambda_a\circ G(\alpha)=\lambda_b\circ G(\beta)$.
Without loss of generality, the objects $\alpha,\beta$ (of $j/F$) are connected by a $t:a\to b$ in $I$ with $F(t)\circ\alpha=\beta$. Then: $$\lambda_b\circ G(\beta)=\lambda_b\circ GF(t)\circ G(\alpha)=\lambda_a\circ G(\alpha)$$As desired.
So, the $(\varsigma_\bullet)$ form a genuine cocone! This induces some $\sigma:\varinjlim_J G\to\varinjlim_I GF$. I claim $\sigma$ is inverse to $\kappa$, and from this we find $\kappa$ to be an isomorphism.
$\sigma\kappa:\varinjlim_I GF\to\varinjlim_I GF$ has every $i$th component equal to: $$\begin{align}\sigma\kappa\circ\lambda_i&=\sigma\circ\mu_{F(i)}\\&=\varsigma_{F(i)}\\&=\lambda_{\iota(F(i))}\circ G(\phi_{F(i)})\\&\overset{\ast\ast}{=}\lambda_i\circ G(\mathrm{Id}_{F(i)})\\&=\lambda_i\end{align}$$By uniqueness, $\sigma\kappa$ is the identity. The step marked $\ast\ast$ uses the fact that the objects $\mathrm{Id}_{F(i)}:F(i)\to F(i)$, $\phi_{F(i)}:F(i)\to F(\iota(F(i))$ are connected in $F(i)/F$ and the previous mini-lemma about connected arrows. Alternatively, there is no loss in generality if you directly choose $\iota(F(i))=i$ and $\phi_{F(i)}:=\mathrm{Id}_{F(i)}$ for all $i\in I$.
And $\kappa\sigma:\varinjlim_J G\to\varinjlim_J G$ has every $j$th component equal to: $$\begin{align}\kappa\sigma\circ\mu_j&=\kappa\circ\varsigma_j\\&=\kappa\circ\lambda_{\iota(j)}\circ G(\phi_j)\\&=\mu_{F(\iota(j))}\circ G(\phi_j)\\&=\mu_j\end{align}$$So $\kappa\sigma$ is also the identity.