follow up from this: Prove that for any $a,b,c,d\in\mathbb{R}$, $[a,b]\sim[c,d]$.
I'm a lot more confident in this proof, I just have one question.
Theorem. If $a<b$ and $c<d$ then $[a, b] \sim[c, d]$
${\textit Proof}$.
Let $f(x)=\frac{d-c}{b-a} x+\frac{c b-a d}{b-a}$.
Then
$$
f(a)=\frac{a(d-c)}{b-a}+\frac{c b-a d}{b-a}=\frac{a d-a c}{b-a}+\frac{c b-a d}{b-a}=\frac{c b-a c}{b-a}=c,
$$
and
$$ f(b)=\frac{b(d-c)}{b-a}+\frac{c b-a d}{b-a}=\frac{b d-b c+c b-a d}{b-a}=\frac{b d-a d}{b-a}=d . $$
Since $f$ is linear, we have $f:[a, b] \rightarrow[c, d]$ with bijectiveness, and therefore $[a, b] \sim[c, d]$. $\square$
From $f(a)=c$, $f(b)=d$ and $f$ is linear, is it valid to infer that $f:[a,b]\to[c,d]$, and that this also has the bijection property?