Let $f:[0,1]\times[0,1]\to\mathbb{R}^2$ be given by $f(x,y)=(x^2+y^2,x^2-y^2)$.
Am I correct in thinking that $f$ has no inverse?
I can show that $f$ is one-to-one but $f$ is not onto since for $(1,2)$ in the codomain, there is no $(x,y)$ in the domain such that $f(x,y)=(1,2)$. So $f$ does not have any inverse.
You are correct; that function, with the specified codomain, has no inverse, because it is not surjective.
Many people will, however, construct something very like an inverse for it, namely, an inverse that maps from $K$ to $[0, 1] \times [0, 1]$, where $K$ is the image of the domain under $f$, and they are likely to informally call this thing an "inverse".