Let $I \subset \mathbb{R}^n$ be non-empty such that $I$ is in general not a vector subspace of $\mathbb{R}^n$, i.e. it is not necessarily closed under addition and multiplication by a scalar. We define a projection from $I$ to $\mathbb{R}^{n-1}$ by $\alpha(\boldsymbol{x}) = (x_1, \ldots, x_{n-1})$, where $\boldsymbol{x} = (x_1, \ldots, x_n), \ \boldsymbol{x} \in I$.
We know that the following property holds for $\alpha$, which follows from the properties of $I$: for $\boldsymbol{x}, \boldsymbol{y} \in I$, if $\alpha(\boldsymbol{x}) = \alpha(\boldsymbol{y})$, then $\boldsymbol{x} = \boldsymbol{y}$.
The paper I read shows that $\dim(I) = n-1$ and thus the Lebesgue measure of $I$ in $\mathbb{R}^n$ is $0$. Its reasoning is that $\alpha$ is an epimorphism since it is a projection and a monomorphism by the aforementioned property, so it is an isomorphism.
My questions are:
- Which definition of an isomorphism is used here at all? (I am mostly familiar with the definition of an isomorphism between two vector spaces).
- Is it true to state that $\alpha$ is an isomorphism because it is a bijection (i.e. leaving out the notions of epimorphism and monomorphism)?
- If no, which definitions of epimorphism and monomorphism are used here and why these two properties imply that $\alpha$ is an isomorphism?
References to definitions and theorems in books are also appreciated very much. Thank you!
You can think of $I$ as the graph of a function $$f:\mathbb{R}^{n-1}\to \mathbb{R},\: (x_1,\ldots,x_{n-1})\to f(x_1,\ldots,x_{n-1}).$$ That is,
$$I=\{(x_1,\ldots,x_n)\in \mathbb{R}^n:x_n=f(x_1,\ldots,x_{n-1})\}.$$ Assuming $f$ is continuous then isomorphism should mean homeomorphism. If $f$ is not continuous, then isomorphism would be just bijection.