Let $\hat{X}$ be the compactification of a Locally compact Hausdorff-space $X$.
Show, that it exists an unique, continuous function $p_{\hat{X}}:\hat{X}\to X^+$, whose restriction on $X$ is the identity.
Hello, I have a question to this task. The function should be obviously
$p_{\hat{X}}(x)=\begin{cases} x\quad\text{for}\,\, x\in X\\ \infty\quad\text{else}\end{cases}$
Then is $p_{\hat{X}}$ restricted on $X$ the identity, by construction. How can I show, that this function is unique. So there does not exist an other function $g$ with the same properties.
And how can I show, that this function is truly continuous?
Assume there exists an other continious function $g:\hat{X}\to X^+$ such that the restriction on $X$ is the identity.
For every $x\in X$ is obviously $p_{\hat{X}}(x)-g(x)=0$ so $p_{\hat{X}}(x)=g(x)$. When $x\in\hat{X}\setminus X$, then I would get
$p_{\hat{X}}(x)-g(x)=\infty-\infty$ what does not make sense at all...
How can I show, that this function is unique? And how can I show that it is continuous?
Thanks in advance.
First of all $p_{\hat{X}}-g(x)$ doesn't make much sense since $X^+$ is only a topological space and not a vector space.
To check that the map is continous you need to check that the preimages of open sets are open set. So you need to specify a topology on both spaces. I guess that $X^+$ is the one point compactification and therefore the topology is just the open sets of $X$ plus sets $V$ which contain $\infty$ s.t. $X^+\setminus V$ is closed and compact in $X$. And I will use that $X$ is embedded in $\hat{X}$, meaning that open sets of $X$ are open in $\hat{X}$.
For the open sets in $U\subset X\subset X^+$ it is clear that the preimage is again $U$ und therefore open in $\hat{X}$. Now let $V$ be a subset of the second form, then $K:=X^+\setminus V$ is closed and compact in $X$. Since preimages behave nicely with respect to complements you have $K=p_{\hat{X}}^{-1}(K)=p_{\hat{X}}^{-1}(X^+\setminus V)=\hat{X}\setminus p_{\hat{X}}^{-1}(V)$. This is closed since it is compact in $X$ and therefore also in $\hat{X}$. You can conclude that $ p_{\hat{X}}^{-1}(V)$ is open in $\hat{X}$.
For uniqueness assume there is another $f:\hat{X}\to X^+$ with some point $x \in \hat{X}\setminus X$ s.t. $f(x)\neq p_{\hat{X}}(x)$. By the construction of $p_{\hat{X}}$ this means that $f(x) \in X$. Choose a relatively compact neighborhood $V$ of $f(x) \in X$, then $U=\hat{X}\setminus \overline{V}$ is a neighborhood of $x$. Put $W=U\cap f^{-1}(V)$. We have that $W$ is nonempty since $x\in W$. Furthermore we have that $W\cap X= \emptyset$ since $f^{-1}(V)=V \cup (\hat{X}\setminus X\cap f^{-1}(V))$ and $U$ and $V$ are disjoint. So we conclude that $\hat{X}\setminus X\cap f^{-1}(V)$ is open and nonempty which contradicts the denseness of $X$ in $\hat{X}$.