show that f is homomorphism?

Question

under the usual Topology on $\mathbb{R}^3$ ,the map $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ defined by $f(x,y,z) =(x+1,y-1,z)$ is

choose the correct option

$a)$ open

$b)$ closed

$c)$ neither open nor closed

$d)$ both open and closed

I thinks option $d)$ will correct because $\mathbb{R}$ is both open and closed

Is its True ??

Any hints/solution will be appreciated

thanks u

Answer 1

Prescribe $g:\mathbb R^3\to\mathbb R^3$ by:$$(x,y,z)\mapsto(x-1,y+1,z)$$

Then $f\circ g=g\circ f=\mathsf{id}_{\mathbb R^3}$.

Further both functions are continuous.

From this it follows that $f$ is a homeomorphism, hence is open and closed.