Munkres topology question section 24 Question.7?

Question

Let $X\subset \Bbb R$ be a topological space $(-\infty , -1) \cup [0 , \infty)$ equipped with the subspace topology on $\mathbb{R}$. Show that the function $f \colon X \to \Bbb R$ defined by $$f(x)=\begin{cases}x+1 &\text{if} \ x <-1 \\ x &\text{if} \ x \geq 0,\end{cases}$$ is order preserving surjective. Furthermore, is $f$ a homeomorphism?

I was reading this article and I got some clues and hints from it but, I'm not getting how to tackle with this problem.

Answer 1

Your function is order preserving because, if $x,y\in X$ and $x<y$, then

1. if $x<y<-1$, then $f(x)<f(y)$, since $f(x)=x+1$ and $f(y)=y+1$;
2. if $0\leqslant x<y$, then $f(x)<f(y)$, since $f(x)=x$ and $f(y)=y$;
3. if $x<-1$ and $0\leqslant y$, then $f(x)<f(y)$ because $x<-1$ and $0\leqslant y$ implias that $y-x>1$, which means that $x+1<1$; in other words, $f(x)<f(y)$.

And it is surjective, because, if $\in\mathbb R$:

1. if $y\geqslant0$ and you take $x=y$, then $f(x)=y$;
2. if $<y0$ and you take $x=y-1$, then $f(x)=x+1=y$.

And it is not a homomorphism, because $X$ is not connected, whereas $\mathbb R$ is.

Answer 2

For $X=(-\infty,1)\cup[0,\infty)$, both $(-\infty,1)$ and $[0,\infty)$ are nonempty, they are open in $X$, and they are disjoint, so $X$ is not connected. And we know homeomorphism preserves connectedness, so $f$ is not homeomorphism.

It is surjective since for $y\geq 0$, we can have $f(y)=y$. For $y<0$, then $y-1<-1$ and $f(y-1)=y-1+1=y$.

Order preserving: For $x_{1},x_{2}\in[0,\infty)$ such that $x_{1}\geq x_{2}$, then of course $f(x_{1})=x_{1}\geq x_{2}=f(x_{2})$.

For $x_{1}\in[0,\infty)$ and $x_{2}\in(-\infty,-1)$, then $f(x_{1})-f(x_{2})=x_{1}-(x_{2}+1)\geq 0$.

For $x_{1},x_{2}\in(-\infty,-1)$, then $f(x_{1})-f(x_{2})=x_{1}-x_{2}\geq 0$.