One nappe of the hyperbola is an embedding. Pollack 1.3.8 The problem asks to check that the map $f:\mathbb{R}^1\to \mathbb{R}^2$ given by $t\mapsto(\cosh(t),\sinh(t))$ is a closed embedding. I tried two different approaches to solve this problem.
First, we can use the fact that
$$f \text{ is a closed embedding if and only if } f(\mathbb{R}^1) \text{ is closed in }\mathbb{R}^2$$
and $f$ is a homeomorphism onto its image.

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*We can see that $f(\mathbb{R}^1)$ is closed since its complement is open.

*Next, we want to show that $f$ is a homeomorphism. If we are going to look at it geometrically, then it is obvious as we can see that the image/preimage of open sets is going to be open. My first question: how to show this more precisely?
I was thinking to define a function $g:f(\mathbb{R}^1)\to \mathbb{R}^1$ given by $g(x,y)=\ln(x+y)$ since
$$(x,y)=(\cosh(t),\sinh(t))\text{, so } x+y=e^t\text{ i.e. }t=\ln(x+y)$$
So, we can see that $f$ and $g$ are continuous functions and inverses of each other i.e. $f$ is a homeomorphism. Does it work? (1)
Second approach is to use the direct definition of the closed embedding. In other words,
$$f\text{ is a closed embedding if and only if } f\text{ is an immersion and the preimage of every compact set is compact.}$$

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*$f$ is an immersion if the differential $df_a$ is injective for all $a\in\mathbb{R}^1$ i.e. the tangent vector is never zero. But, we can see that $df_a=\begin{bmatrix}\sinh(a)\\ \cosh(a)\end{bmatrix}$, so $|df_a|\neq0$

*To show that the preimage of a compact set is compact, I used the following. If we take any compact set $A$ in $\mathbb{R}^2$, then we always can find a closed ball center at the origin that will contain $A$. That closed ball will intersect $f(\mathbb{R}^1)$ at some point $(x_0,y_0)$ i.e. at some point $t_0\in \mathbb{R}^1$ as $(x_0,y_0)\in f(\mathbb{R}^1)$. Then we can see that
$$f^{-1}(A\cap f(\mathbb{R}^1))\subset[-t_0,t_0]$$
Since $A\cap f(\mathbb{R}^1)$ is closed, $f$ is continuous, and $[-t_0,t_0]$ is compact, then $f^{-1}(A\cap f(\mathbb{R}^1))$ is compact as the closed subset of the compact set. Does it work? (2)
 A: First approach: You claimed, but did not prove, that the complement of $f(\Bbb R)$ is open. It is easier to prove that $f(\Bbb R)$ is closed. In fact,$$f(\Bbb R)=\{(x,y)\in\Bbb R^2\mid x\geqslant0\}\cap\{(x,y)\in\Bbb R^2\mid x^2-y^2=1\}.\tag1$$The first set is closed, since it is $\varphi^{-1}\bigl([0,\infty)\bigr)$ with $\varphi(x,y)=x$ and $\varphi$ is continuous; and the second set is closed, since it is $\psi^{-1}\bigl(\{1\}\bigr)$  and $\psi$ is continuous. So, $(1)$ is closed.
And, in order to prove that $f$ is a homeomorphism onto its image, what you dis is correct, but it is simpler to use the fac that $\sinh$ is a homeomorphism (from $\Bbb R$ onto itself) and that $(x,y)\mapsto\sinh^{-1}(y)$ is the inverse of $f$.
Second approach: Use the fact that every compact subset of $\Bbb R^2$ is contained in some set of the form $\{(x,y)\in\Bbb R^2\mid x\leqslant M\}$, for some $M\geqslant1$, and that, if $t_0\in[0,\infty)$ is such that $\cosh(t_0)=M$, then $f^{-1}(A)\subset f^{-1}(M)=[-t_0,t_0]$.
