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I have to show hat the following two figures are homeomorphic to each other. Let the first and second figures be denoted as $X,Y$ respectively.

$X$ is is obtained by sewing together three twisted strips of paper to two circular discs of paper and $Y$ is obtained by sewing together two long strips of paper as shown in the figure.

My attempt: Since a double twisted Mobius band is homeomorphic to a cylinder, figure $X$ is homeomorphic to "A Mobius band with a small disk removed from it's interior" ( I hope it is fine. If not, you can give your argument why is it not right and then proceed in another way.) Also we can show $Y$ is homeomorphic to "A torus with a small disk removed from it". Now we know that if $f:Y\rightarrow X$ is a homeomorphism then $f(\partial Y)=\partial X$. Obviously $f|_{\partial Y}$ is continuous. But here as per my description $\partial X$ has two components while $\partial Y$ has only one component, hence they can't be homeomorphic to each other.

So please find the fault in my argument and give me a hint for proving this! If these are not homeomorphic, then also give me some hint for disproving this(though I've disproved it, i need to hear if there is any other way we can think about this).

Also note that, both have same homotopy type as both of them deform retract to figure $\infty$ , so there is a chance of these figures to be homeomorphic.

P.S. Well, you can provide me an intuitive answer also!

enter image description here

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I don't think the figure $X$ is a Mobius band with a disc removed. For one thing, it has only one boundary component (just by tracing your finger around the arcs and seeing that cover everything). Secondly, it is orientable, since you always pass through an even number of twisted bands when you travel by any closed curve.

If you're able to use the classification of surfaces theorem, then you could verify the two are homeomorphic by calculating the Euler characteristic (which is $-1$ for both), number of boundary components ($1$ for both), and the fact both are orientable.

For a more direct way to see they are homeomorphic, we start by looking at the figure $Y$. It is an annulus with a strip glued onto it. So we want to describe $X$ in the same way. Now if we cut one of the twists from the figure $X$, we are left with a band with two twists in it, which is homeomorphic to an annulus. So $X$ is in fact homeomorphic to an annulus with a band glued to it. There are two ways to glue on a band, one which results in an orientable surface and one which results in a non-orientable surface, and we know $X$ is orientable. So the surface $X$ is homeomorphic to the orientable surface where we glue a band to an annulus, which is also homeomorphic to $Y$.

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    well, i don't have clear idea about orientability of spaces! I also don't know "classification of surfaces theorem". What you are saying, i understood. But, is it not true that "doubly twisted mobius band is homeomorphic to cylinder"? then where did my argument go wrong? – 2017-02-28
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    I would like to ask if you can provide a more basic way – 2017-02-28
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    It is true that a doubly twisted Mobius band is homeomorphic to the cylinder, but note that the figure X is not a triply twisted band: the bands are glued together in a different way. If you go round a closed curve in a triply twisted band, you pass through all three twists, but in the figure there are closed curves that only pass through two twists. – 2017-02-28
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    A Mobius band with a disc removed would have two boundary components, but we can see from the figure that it has only one boundary component. – 2017-02-28
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    I have added what I hope is an intuitive way to see that $X$ and $Y$ are homeomorphic. – 2017-02-28
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    How do we know that $X$ is orientable? – 2017-02-28
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    One way to think about orientability in this case is by imagining an arrow pointing out of the surface pushing that arrow around the surface. If the surface is non-orientable, then it is possible to come back to the same point with the arrow pointing the opposite way (eg. in the Mobius band). With an orientable surface, the arrow will always face the same way when we come back, which is the case with $X$. – 2017-02-28