According to Lee's book "Introduction to topological manifolds" (also I believe this is the standard way to define it) a topological manifold of dimension $n$ is a second countable Hausdorff space $\mathcal{M}$ such that for every $p \in \mathcal{M}$ there exists an open neighbourhood
$U$ of $p$ and a homeomorphism
\begin{equation}
\varphi\colon U \to \varphi(U)\subseteq \mathbb{R}^n.
\end{equation}
Let us assume $\mathcal{M}$ is such an $n$-dimensional topological manifold and take any $p\in \mathcal{M}$. We know based on the definition above that there exists an open neighbourhood $U$ of $p$ and a homeomorphism $\varphi$ into a subset of $\mathbb{R}^n$.
Now define the map
\begin{align}
\tilde{\varphi}\colon U &\to \tilde{\varphi}(U)\\
x &\mapsto \tilde{\varphi}(x):= (\varphi(x),0).
\end{align}
Then $\tilde{\varphi}$ is again a homeomorphism and further $\tilde{\varphi}(U) = \varphi(U)\times \{0\}\subseteq\mathbb{R}^{n+1}$. Thus we can construct a homeomorphism
into a subset of $\mathbb{R}^{n+1}$. The same manifold $\mathcal{M}$ could be therefore also interpreted to be a topological manifold of dimension $n+1$.
Where is my mistake?
I suppose you could just get rid of all such redundant "$0$" components first and then define the dimension in a sense to be the "minimal" one. But I have not found any comment
on this yet.
Best Answer
In the definition, it's required that $\phi(U)$ is open in $\mathbb R^n$, then $\phi(U)\times\{0\}$ is not open in $\mathbb R^{n+1}$.
Now a connected manifold cannot have two distinct dimensions, since a non-empty open subset of $\mathbb R^n$ can never be homeomorphic to one in $\mathbb R^m$ unless $m=n$. This is a non-trivial but standard result from algebraic topology.