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).

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    $\begingroup$ Yes: take any bijection from $\mathbb{R}$ to $\mathbb{R}^2$, and choose the topology that makes it a homeomorphism! $\endgroup$ Mar 4 at 13:46
  • $\begingroup$ 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. $\endgroup$ Mar 4 at 13:51
  • $\begingroup$ @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 $\endgroup$
    – yi li
    Mar 4 at 13:51
  • $\begingroup$ @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 $M$? $\endgroup$
    – yi li
    Mar 4 at 13:59
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    $\begingroup$ 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. $\endgroup$
    – Dan Rust
    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 $(N,\tau_N)$.


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