# Can a set equipped with two different topology makes into topological manifold with different dimension?

Let $$M$$ equipped with two different topology $$\tau_1$$ and $$\tau_2$$ .Is it possible that $$\tau_1$$ and $$\tau_2$$ makes $$M$$ into topological manifold with different dimension ?

There is a bit difference between this question and topological invariance of dimension (which states that dimension of topological manifold is invariance under homeomorphism).

• Yes: take any bijection from $\mathbb{R}$ to $\mathbb{R}^2$, and choose the topology that makes it a homeomorphism! Mar 4 at 13:46
• In fact, you can take a bijection between any two topolocial manifolds of positive dimension (they all have the cardinality of $\mathbb R$) and declare it to be a homeomorphism. Mar 4 at 13:51
• @Johnny El Curvas Do you mean taking the space filling curve with topology on that image of curve open if and only if inverse image is open in $\Bbb{R}$?then the topology makes the space 1-dimension manifold under this topology Mar 4 at 13:51
• @Andreas Cap Do you mean first give the bijective map $f:M\to N$ ,then fixed the topology on $N$ and pull back the structure on $N$ to $M$? Mar 4 at 13:59
• Space filling curves are not bijective in general. Johnny El Curvas just said a bijection. Obviously a bijection exists from $\mathbb{R}$ to $\mathbb{R}^2$ (that is elementary set theory), so just use that bijection and then define your open sets in $\mathbb{R}$ to be precisely the preimages of the open sets in $\mathbb{R}^2$ under this bijection. Mar 4 at 14:34

All nonempty manifolds of any positive dimension have the cardinality $$\mathfrak c$$ of the continuum because they are covered by countably many open sets homeomorphic to $$\mathbb R^n$$ for some $$n > 0$$. Since all $$\mathbb R^n$$ have the cardinality $$\mathfrak c$$ (you do not need topology for this fact!), our claim follows.
Thus for any two nonempty manifolds $$(M,\tau_M), (N,\tau_N)$$ of positive dimension there exists a bijection $$f : M \to N$$. Now consider the topology $$f^{-1}(\tau_N) = \{f^{-1}(U) \mid U \in \tau_N\}$$. Then $$(M,f^{-1}(\tau_N))$$ is homeomorphic to $$(N,\tau_N)$$.