# Find continuous functions $f,g$ such that $g\circ f$ is closed and continuous but neither $g$ nor $f$ is closed map.

Find continuous functions $$f,g$$ such that $$g\circ f$$ is closed and continuous but neither $$g$$ nor $$f$$ is closed map.

Find continuous functions $$f,g$$ such that $$g\circ f$$ is open and continuous but neither $$g$$ nor $$f$$ is open map. Consider the spaces to be $$\Bbb R$$

I took $$f(x)=$$ $$\begin{cases} (x-1) &-1\le x\le 1\\0 & x>1,x<-1\end{cases}$$ and

$$g(x)=$$ $$\begin{cases} (x-1) &x>1,x<-1\\0 & -1\le x\le 1\end{cases}$$

Though $$g\circ f=0$$ which is closed and $$f$$ is not closed since it takes $$(-0.5,0.5)$$ to $$0$$ but here $$g$$ is closed

I want examples where $$g,f$$ are not closed both. How to make $$g$$ not closed?

How to answer the second question?

• A little off topic question, but why the measure theory tag? Jan 27, 2019 at 7:36
• Does the topology have to be the usual one in $\mathbb{R}$ ? Jan 27, 2019 at 13:19
• Hint: If $g$ is non-closed, and it is constant on some non-closed interval, then the restriction to that interval is closed. Jan 27, 2019 at 14:21
• Each of your functions $f$ and $g$ is discontinuous at $-1$. Feb 2, 2019 at 21:21

Find continuous functions $$f,g$$ such that $$g\circ f$$ is closed and continuous but neither $$g$$ nor $$f$$ is closed map.
We shall following a hint by user87690. First we put $$f(x)=e^x$$. Since $$f(\Bbb R)=(0,\infty)$$, the map $$f$$ is not closed. Now let $$g(x)$$ equals $$1$$, if $$x\ge -1$$ and $$-1/x$$, otherwise. Since $$g(\Bbb R)=(0, 1]$$, the map $$g$$ is not closed. But $$gf(x)=1$$, so $$gf(A)=\{1\}$$ for each non-empty closed subset $$A$$ of $$\Bbb R$$.
Find continuous functions $$f,g$$ such that $$g\circ f$$ is open and continuous but neither $$g$$ nor $$f$$ is open map.
There are no such functions. Indeed, assume to the contrary that the function $$gf:\Bbb R\to\Bbb R$$ is open. First we show that the function $$gf$$ in injective. Suppose to the contrary that there exists $$x,x’\in\Bbb R$$ such that $$x and $$gf(x)=gf(x’)=y$$. Since $$gf$$ is a continuous map, an image $$gf([x,x’])$$ of the segment $$[x,x’]$$ is compact. Therefore there exist numbers $$m=\min gf([x,x’])$$ and $$M=\max gf([x,x’])$$. If $$m then an image $$gf((x,x’))$$ of an open set $$(x,x’)$$ equals to $$[m,M]$$ or to $$[m,M)$$, but neither of these sets is open, a contradiction. Similarly we obtain a contradiction if $$M>y$$. If $$m=M=y$$ then $$gf((x,x’))=\{y\}$$, a contradiction again. Thus $$gf(x)\ne gf(x’)$$. Thefore the function $$f:\Bbb R\to\Bbb R$$ in injective too. By invariance of domain, $$f$$ is an open map.