Compactifications $Y_1\geq Y_2$

Question

Let $i_1:X\rightarrow Y_1$, $i_2:X\rightarrow Y_2$ be compactifications of a space $X$. Show that:

1. If $Y_1\geq Y_2$, then there exists only one map $g:Y_1\rightarrow Y_2$ satisfying $i_2=gi_1$; moreover, $g$ is onto.

2. $Y_1\geq Y_2$ and $Y_2\geq Y_1$ if and only if the map $g:Y_1\rightarrow Y_2$ is a homeomorphism.

Here's what I know: $Y_1\geq Y_2$ means that $g$ exists as defined above, and $Y_2\geq Y_1$ means there exists a continuous function $h$ such that $i_1=hi_2$. I assume that in the question, $g^{-1}=h$.

1. Assume there exist two functions, $g_1$ and $g_2$ such that $i_2=g_1i_1$ and $i_2=g_2i_1$. Somehow I have to use compactification to make a contradiction.

2. By definition, $g$ and $g^{-1}$ are continuous. So it remains to be shown that $g$ is a bijection. Without an explicit definition of $g$, I don't know how to show this.

Answer 1

If $g\circ i_1=i_2$ and $g^\star\circ i_1=i_2$, then $g=g^\star$ because $g$ and $g^\star$ are continuous, their restrictions to $i_1(X)$ are identical and $i_1(X)$ is a dense subset of $Y_1$.

If there is a continuous map $h\colon Y_2\longrightarrow Y_1$ such that $h\circ i_2=i_1$, then $h\circ g\circ i_1=i_1$. And, of course, $\operatorname{id}_{Y_1}\circ i_1=i_1$. So, by the same reason as above, $h\circ g=\operatorname{id}_{Y_1}$. For the same reason, $g\circ h=\operatorname{id}_{Y_2}$. Therefore, $g$ is a homeomorphism.

Answer 2

That $i_1: X \to Y_1$ is a compactification implies that $i_1[X]$ is dense in $Y_1$.

Suppose we had two maps $g_1,g_2: Y_1 \to Y_2$ obeying $i_2 = g_i \circ i_1$, for $i=1,2$.

Then $g_1$ and $g_2$ coincide on the dense subset $i_1[X]$ of $Y_2$ and as $Y_2$ is Hausdorff, $g_1 = g_2$.

Now suppose $Y_1 \ge Y_2$, as shown by $g: Y_1 \to Y_2$ with $g \circ i_1 =i_2$ and also $Y_2 \ge Y_1$, witnessed by $h: Y_2 \to Y_1$ with $h\circ i_2 = i_1$.

Now $h \circ g: Y_1 \to Y_1$ and $$(h \circ g) \circ i_1 =h \circ (g \circ i_1) = h \circ i_2 =i_1$$.

Hence $Y_1 \to Y_1$ is witnesses by $(h \circ g)$ The function $\text{id}_{Y_1}$ also witnesses that $Y_1 \to Y_1$ and the first part thus shows that $h \circ g = \text{id}_{Y_1}$ by unicity.

The argument with swapped order shows that $g \circ h = \text{id}_{Y_2}$ as well and so $h,g$ are each other's inverse (both ways) and so are homeomorphisms.