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What do we mean mathematically when we say that a coffee mug is topologically equivalent to a donut ? How do we decide which shapes can be transferred into which shapes ? I guess people would say "Homeomorphism " is the key to such questions. So , is it so that if we dan find a homeomorphism from one shape to another then we can say that shape $\ A$ can be transformed into shape$\ B$ . But , again homeomorphism depends on the topology that is chosen . If , I declare each and every subset of the power set of the set of points in both shapes $\ A$ and$\ B$ to be open , then define a mapping such that for every point in $\ A$there lies a unique point in $\ B$ and vice versa , then this mapping can be a homeomorphism .(Correct me , if I am wrong ). And by this logic all the shapes are topologically equivalent . But , I can smell something is fishy around here . Could someone help me in understanding this ? What is an example of a shape that is not topologically equivalent to a sphere? What is an example of a shape that can not be obtained from a square ?

Thanks ahead!

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  • $\begingroup$ Not homeomorphism but homotopy. Homotopy equivalence between (and the types of) topological spaces, manifolds, surfaces, etc. It's the precise meaning of "deformable". $\endgroup$ – Andrew Miloradovsky May 29 '16 at 19:01
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If $X$ and $Y$ are topological spaces of the same cardinality, and one chooses $\mathcal{P}(X)$ and $\mathcal{P}(Y)$ as topologies for $X$ and $Y$ respectively, then indeed every bijection $f: X \rightarrow Y$ is a homeomorphism. So in that sense, choosing this topology for all surfaces makes all surfaces homeomorphic(since in general the cardinality of a surface is the cardinality of $\mathbb{R}$). However, when people talk about the torus for example, they mean the torus with the standard subspace topology inherited by $\mathbb{R}^3$.

So when people say the torus is not homeomorphic to the sphere they mean that there is no bijective function $\varphi: S^2 \rightarrow T $ with the property that $U \subset S^2$ is open with respect to the subspace toplogy of $S^2$ in $\mathbb{R}^3$ if and only if $\varphi(U) \subset T$ is open with respect to the subspace topology of $T$ in $\mathbb{R}^3$. There is not only a bijection between the two topological spaces, but the bijection is compatible with the topological structure. Just think of a topology as a way of saying in what way the points of the set you start with are glued together. Then there is a homeomorphism $\varphi: X \rightarrow Y$ between two spaces $X$ and $Y$ if you can assign to each point $ x \in X$ a point $\varphi(x) \in Y$ such that sets of points are glued together in $X$ if and only their images are glued together in $Y$.

I hope this helps.

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  • $\begingroup$ :Sorry for replying late . But what are the techniques to prove that two spaces are homeomorphic ? $\endgroup$ – itp dusra Jun 19 '16 at 17:47
  • $\begingroup$ As far as I know, showing two arbitrary topological spaces are homeomorphic can be quite difficult, and I do not know of standard techniques for doing so, other than constructing a homeomorphism. However, there are lots of ways to show two topological spaces , say $X$ and $Y$, are not homeomorphic. One typical way of doing this is showing $X$ has some 'topological property' $Y$ does not have. For example, if $X$ and $Y$ are path connected, we can, up to isomorphism, attach groups $\pi_1(X)$ and $\pi_1(Y)$ to $X$ and $Y$ respectively. $\endgroup$ – M. Van Jun 19 '16 at 19:32
  • $\begingroup$ We call $\pi_1(X)$ the fundamental group of $X$. It basically consists of loops(that is, functions $f: [0,1] \rightarrow X$) based at some point $x_0 \in X$. Look up the construction of the fundamental group for more details. Now if $\pi_1(X)$ is not isomorphic to $\pi_1(Y)$ as a group, then $X$ and $Y$ cannot be homeomorphic. This method is useful because groups are generally better understood than topological spaces, so one can use group theory to understand topology. $\endgroup$ – M. Van Jun 19 '16 at 19:35
  • $\begingroup$ Thanks a lot . I needa learn more to get deeper into this . $\endgroup$ – itp dusra Jun 22 '16 at 18:39

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