[For convenience, I will pretend in this answer that any compactification actually contains $X$ as a subspace, so I don't have to constantly be writing down the embedding maps.]
No. For instance, you can define a compactification $Y=X\cup\{\infty\}$ where the only neighborhood of $\infty$ is the entire space (and every open subset of $X$ remains open). There will not exist any morphism from this compactification to $X^*$ unless the topology on $X^*$ happens to be the same as the topology on $Y$ (i.e., the only compact closed subset of $X$ is the empty set; given your assumption that $X$ is KC and noncompact, this is impossible!).
A separate issue is that $X$ is open in $X^*$, so if you have some other compactification in which $X$ is not open, you cannot expect it to have a morphism to $X^*$. There are also uniqueness issues--a morphism to $X^*$ does not need to send all the new points to $\infty$ (for instance, if $X$ is uncountable with the cocountable topology, you could let $X'$ be $X$ together with one more point with the cocountable topology and let $Y$ be the 1-point compactification of $X'$, and then the new point of $X'$ can map to anywhere in $X^*$ and the map will still be continuous). With non-Hausdorff spaces, a continuous map is not determined by its values on a dense subset, so generally it will be very hard to get any sort of uniqueness property like this without stronger hypotheses.
If you restrict your definition of "compactification" to require $Y$ to also be KC and that $X$ is open in $Y$, then it is true that $X^*$ is the terminal compactification (assuming $X^*$ is a compactification at all by this definition--it won't always be KC). These hypotheses make it trivial to check that the map $Y\to X^*$ sending every new point to $\infty$ is continuous (the hypothesis that $X$ is open in $Y$ gives continuity at points of $X$, and the hypothesis that $Y$ is KC gives continuity at new points).
For uniqueness, suppose $h:Y\to X^*$ is a morphism of compactifications, and let $A=X\cup h^{-1}(\{\infty\})\subseteq Y$. Then I claim $A$ is compact. To prove this, note that $h^{-1}(\{\infty\})$ is closed in $Y$ and hence compact, so it suffices to show any ultrafilter $F$ on $X$ has a limit in $A$. By compactness of $Y$, $F$ has a limit $y\in Y$; if $y\in A$ we're done, so we may assume $y\not\in A$. In that case $h(y)\neq\infty$, so it is a point of $X$, and then since $h$ is the identity on $X$, $F$ must converge to $h(y)$ in $X$. Thus $h(y)$ is a limit of $F$ in $A$.
Thus since $Y$ is KC, $A$ is closed in $Y$. Since $A$ contains $X$ and $X$ is dense in $Y$, this means $A=Y$. Thus $h$ must map every point of $Y\setminus X$ to $\infty$.
$\newcommand{\rmod}{\mathsf{AMod}}\newcommand{\modr}{\mathsf{ModA}}\newcommand{\ab}{\mathsf{Ab}}\newcommand{\C}{\mathsf{C}}$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:
- The discrete diagram to $x$ and $y$ has a cocone $x\to z,y\to z$, so in particular there is an arrow $(x,y)\to\Delta(z)$ in $I\times I$ and the coslice category is nonempty
- For every $(f,g):(x,y)\to(i,i)$ and $(u,v):(x,y)\to(j,j)$ in $I\times I$ we may construct a cocone under $f,u:x\to i,j$ to obtain some $\alpha:i\to a$, $\beta:j\to a$ with $\alpha f=\beta u$. We may similarly find a $b$ and some $\gamma:i\to b$, $\delta:j\to b$ such that $\gamma g=\delta v$. Then: $$(i,i)\overset{(\alpha,\gamma)}{\longrightarrow}(a,b)\overset{(\beta,\delta)}{\longleftarrow}(j,j)$$Will connect $(f,g,\Delta(i))$ to $(u,v,\Delta(j)$ in the coslice category.
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.
Best Answer
I think the idea is good; let me comment on how I would think about it:
In order for the inverse limit $\varprojlim_{i \in I} X_i$ to be defined, you need $I$ to have the structure of a category, and you need $i \mapsto X_i$ to have the structure of a functor. You do this by making $I$ into a preorder when you define the "$\leq"$ relation, but to be super-pedantic, you didn't explicitly check that the relation $\leq$ is reflexive and transitive. Then you do check that $i \mapsto X_i$ is a functor with the action $(i\leq j) \mapsto \iota_{ij}$ on morphisms. To be super-pedantic again, you never explicitly said that $X$ was to be a functor to the category $Set$ (rather than to some other category); I suppose this was understood, but see the second bullet below.
You then define a cone on the functor $X: I \to Set$, with vertex $\cap_{i \in I} X_i$ and legs $\iota_i$, and verify that this is indeed a cone. Great.
Next you verify the universal property, which I think looks good modulo the one significant issue of your post which I come to next.
The only real issue is the following. In the hypotheses, you never stipulated that the family of sets $(X_i)_{i \in I}$ be downward directed in the sense that for every $i,j \in I$ there exists $k$ with $X_k \subseteq X_i \cap X_j$. But you use this assumption in your proof of the universal property, when you produce the index $k$.
So what you've shown is that if $(X_i)_{i \in I}$ is a family of subsets of $X$ which is downward-directed, then $\varprojlim_{i \in I} X_i = \cap_{i \in I} X_i$. This is good. Without the hypothesis that $(X_i)_{i \in I}$ be downward-directed, the statement would be false in general. For example, suppose that $I = \{1,2\}$ and neither $X_1$ nor $X_2$ is contained in the other. Then $I$ is a discrete poset with the $\leq$ order you define, and $\varprojlim_{i \in I} = X_1 \times X_2$, which of course is typically different from $X_1 \cap X_2$.
Some related notes:
In the usage I'm familiar with, the terms "inverse limit" and "projective limit" are ambiguous. Sometimes they mean (1) "limit indexed by a downward-directed poset" (or more generally, "limit indexed by a co-filtered category"), but sometimes they mean more generally (2) "limit indexed by an arbitrary category". I think that at least nowadays most people tend to just say "limit" when they mean (2), so that the usage (1) is more common, but it's something to watch out for.
In general the limit $\varprojlim_{i \in I} X_i$ can change depending what category you regard the assignment $i \mapsto X_i$ as taking values in. Let's go back to the example where $I = \{1,2\}$ is discrete. If you regard the assignment $i \mapsto X_i$ as taking values in the category $Set$, then as remarked above, you get $\varprojlim_{i \in I} X_i = X_1 \times X_2$, the cartesian product. But instead, you might regard $i \mapsto X_i$ as a functor $I \to Set/ X$, where $Set / X$ is the slice category, i.e. the category where an object is a set $Y$ equipped with a map $Y \to X$, and a morphism is a map $Y \to Y'$ making the obvious diagram commute. If you do this, then you do end up getting $\varprojlim_{i \in I} X_i = X_1 \cap X_2$. Similarly, if you regard $i \mapsto X_i$ as taking values in the poset $Sub(X)$ of subsets of $X$ and inclusions (note that this is a full subcategory of $Set/X$), then you again get $\varprojlim_{i \in I} X_i = X_1 \cap X_2$.
Similarly, the limit can change depending on what we regard the morphisms of $I$ as being, so it's good that you were explicit about this. An exception is when we compute limits in a category which is a preorder. In this case, $\varprojlim_{i \in I} X_i$ is the greatest lower bound of the $X_i$'s (either existing if the other does), and it doesn't matter what category structure you regard $I$ as having for this to be true.