Proposition

Let $U \subset \mathbb{R}^n$ be an open subset. Let $f: U \rightarrow \mathbb{R}^k$ be any continuous map. Let $f_1, \cdots, f_k$ denote each coordinate function of $f$.

Let $F: U \rightarrow \mathbb{R}^{n + k}$ denote the mapping $(x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_n, f_1(x), \cdots, f_k(x))$. Let $M = F(U)$.

• $F$ is a homeomorphism between $U$ and $M$.
• $M$ is an $n$-manifold in $\mathbb{R}^{n + k}$.

Solution

We will restrict $F$’s range to $U$.

• Continuous?
  • Each component function is continuous, so $F$ is continuous.
• Injective?
  • If $f(x_1, \cdots, x_n) = f(y_1, \cdots, y_n)$, then $(x_1, \cdots, x_n) = (y_1, \cdots, y_n)$.
• Surjective?
  • $M = F(U)$, so $F$ is surjective.
• Continuous inverse?
  • The inverse is $m \mapsto (\pi_1(m), \pi_2(m), \cdots, \pi_n(m))$ where each $\pi_i$ is the projection of the $i$th coordinate. Since each component function is continuous, the inverse is continuous.

Therefore, $F$ is indeed a homeomorphism.

Since $M$ is a subspace of $\mathbb{R}^{n + k}$, it is Hausdorff and second countable. Let $x \in M$. Then $M$ is a neighborhood of $x$ that is homeomorphic to $U$, an open subset of $\mathbb{R}^n$. Therefore, $M$ is locally Euclidean.

Hence, $M$ is an $n$-manifold.