Prove that the equivalence of 2 knots is an equivalence relation. The definition of equivalence of 2 knots according to Richard H. Crowell and Ralph H. Fox, edition 1963, is:
Assume that $K_{1}$,$K_{2}$ are 2 knots in $\mathbb{R^3}$, then they are equivalent , denoted by $K_{1} \cong K_{2}$, iff $\exists f: \mathbb{R^3} \rightarrow \mathbb{R^3}$, where $f$ is a homeomorphism and such that $f(K_{1}) = K_{2}.$
My Thoughts:
I know that to prove that a relation is an equivalence relation, I have to prove that it is reflexive, symmetric and transitive.
Also, I know that $f$ is a homeomorphism iff $f$ is bijective and continuous and $f^{-1}$ is continuous. and that this leads to $\mathbb{R^3}$ is homeomorphic to $\mathbb{R^3}$, which is trivially true, so what is the benefit of the $f$ function?
Now, for proving that the given relation is reflexive I will take $f =$ identity function , which is a homeomorphism and $f(K_{1}) = K_{1}.$
For proving that it is symmetric,I know the definition which is: if $K_{1} \cong K_{2}$ then $K_{2} \cong K_{1}$, but I do not know how to prove this, could anyone help me in this please?
For proving that it is transitive: I know the definition which is: if $K_{1} \cong K_{2}$ and if $K_{2} \cong K_{3}$ then $K_{1} \cong K_{3}$, and I know that this requires me to prove that if $f(K_{1}) = K_{2}$ and $f(K_{2}) = K_{3}$, then $f(K_{1}) = K_{3}$, but I do not know how to conclude this, could anyone help me please?         
 A: If $K_1\simeq K_2$, then there's a homeomorphism $f$ of $\Bbb R^3$ taking $K_1$ to $K_2$. Then consider $f^{-1}$ for $K_2\simeq K_1$.
For transitivity, we have an $f$ taking $K_1$ to $K_2$ and perhaps another $g$ taking $K_2$ to $K_3$. Just compose them. 
A: The definition says that two knots are equivalent if there exists a homeomorphism $f \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$, such that the restriction of $f$ yields a homeomorphism between the knots. The benefit is exactly that. We have a homeomorphism between the knots which can be lifted to/comes from the entire space where the knots live.
Assume that $K_1 \cong K_2$. This means that we have a homeomorphism $f \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$, such that $f(K_1) = K_2$. We now need to find a homeomorphism $g \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$, such that $g(K_2) = K_1$. How about $g = f^{-1}$? 
For the transitivity just compose the maps that you are assuming to have. By assumption we have $f \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$, such that $f(K_1) = K_2$ and $g \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$, such that $g(K_2) = K_3$. This means we can just consider $g \circ f$ which is a homeomorphism then and by the definition of $f$ and $g$ we have $g(f(K_1)) = K_3$. That means that $K_1$ and $K_3$ are equivalent.
