Definition (0)
Let be $X$ a topological space. So a pair $(h,K)$ is a compactification of $X$ if $K$ is a compact space and if $h:X\rightarrow K$ is an embedding of $X$ in $K$ such that $h[X]$ is dense in $K$. Moreover a compactification $(h,K)$ of $X$ is a $T_2$ compactification if $K$ is a Hausdorf -compact- space.
Definition (1)
If $(h_1,K_1)$ and $(h_2,K_2)$ are two compactifications of some topological space $X$ we say that $(h_1,K_1)\preccurlyeq(h_2,K_2)$ if there exist a continuous function $p$ such that $p\circ h_2=h_1$ and in particular if $p$ is an homeomorphism we say that the two compactifications are equivalent.
Statemet (2)
The relation $\preccurlyeq$ is an order for Hausdorff compactifications.
Proof. If $(h,K)$ is a compactification of some space $X$ then the identity $\text{id}_K$ on $K$ is an homeomorphism such that $\text{id}_K\circ h=h$ and so $(h,K)\preccurlyeq(h,K)$.
If $(h_1,K_1)$ and $(h_2,K_2)$ are two compactifications of some space $X$ such that $(h_1,K_1)\preccurlyeq(h_2,K_2)$ and $(h_2,K_2)\preccurlyeq(h_1,K_1)$ then there exist two continuous functions $p:K_2\rightarrow K_1$ and $q:K_1\rightarrow K_2$ such that $p\circ h_2=h_1$ and $q\circ h_1=h_2$, or rather $(p\circ q)\circ h_1=h_1$ and $(q\circ p)\circ h_2=h_2$; thus $p\circ q=\text{id}_{h_1[X]}$ and $q\circ p=\text{id}_{h_2[X]}$ and so $p$ and $q$ are continuous and bjective -both have a right and left inverse- and such that $p=q^{_1}$ and $q=p^{-1}$ and so it is clear that $K_1$ and $K_2$ are homeomorphic.
Finally if $(h_1, K_1),(h_2,K_2)$ and $(h_3,K_3)$ are three compactification of some space $X$ such that $(h_1,K_1)\preccurlyeq(h_2,K_2)$ and $(h_2,K_2)\preccurlyeq(h_3,K_3)$ then there exist two continuous functions $p:K_2\rightarrow K_1$ and $q:K_3\rightarrow K_2$ such that $p\circ h_2=h_1$ and $q\circ h_3=h_2$ and so $p\circ q:K_3\rightarrow K_2$ is a continuous function such that $(p\circ q)\circ h_3=p\circ(q\circ h_3)=p\circ h_2=h_1$ and so $(h_1,K_1)\preccurlyeq(h_3,K_3)$.
So I ask if the proof is correct. In particular I point out that to prove that the relation $\preccurlyeq$ is antisymmetric it is necessary to prove that $p\circ q=\text{id}_{K_1}$ and $q\circ p=\text{id}_{K_2}$ and not that $p\circ q=\text{id}_{h_1[X]}$ and $q\circ p=\text{id}_{h_2[X]}$ as I did: so in particular I ask to prove that $p\circ q=\text{id}_{K_1}$ and $q\circ p=\text{id}_{K_2}$ -this is really important!!! Could someone help me, please?
Best Answer
Transitivity is fine, that writes itself, an reflexivity by using the identity too.
Antisymmetry (modulo equivalence): If $(h_1,K_1) \preceq (h_2,K_2)$ we have $p: K_2 \to K_1$ with $p \circ h_2 = h_1$ and if also $(h_2,K_2) \preceq (h_1,K_1)$ we have $q: K_1 \to K_2$ such that $q \circ h_1 = h_2$.
It then indeed follows that $(p \circ q) \circ h_1 = h_1$ and this only says that $p \circ q = \text{id}_{h_1[X]}$ and then we can only conclude that $p \circ q= \text{id}_{K_1}$ if we know $K_1$ is Hausdorff (and using the fact that $h_1[X]$ is dense in $K_1$). So in the case of Hausdorff compactifications we can conclude that $p \circ q = \text{id}_{K_1}$, otherwise this need not hold; at least we cannot use a theorem to conclude it.
Similarly if also $K_2$ is Hausdorff $q \circ p = \text{id}_{K_2}$ via the same route. And then $K_1 \simeq K_2$ with embedding-preservation, so $(h_1,K_1)$ and $(h_2, K_2)$ are equivalent embeddings, not equal, so true antisymmetry is impossible here, only on the equivalence class (literally a class too) of compactifications.
Your conclusion that always $p$ and $q$ are mutual inverses seems unlikely to hold, but in practice this order is only considered for Hausdorff compactifications (they have the nicest theory by far) and it's not a practical issue. There can very well be counterexamples for non-compact compactifications, I feel. What I have seen done in that case though (no reference) is to consider $\preceq$ a pre-order only and define equivalent compactifications by the mutual $\preceq$-relation; this is always an equivalence relation, but you don't get that $K_1$ and $K_2$ are necessarily homeomorphic then.