Is the Klein bottle homeomorphic to the union of two Mobius bands attached along boundary circle? Question: Determine whether the Klein bottle is homeomorphic to the union of two Mobius bands attached along their boundary circles.

The Klein bottle is the quotient space
$$
K=I^2 /{\sim}, \quad (x,0)\sim(x,1), \; (0,y)\sim(1,1-y), \; \forall x,y\in I
$$
The Möbius band is the quotient space
$$
M=I^2 /{\sim}, \quad (0,y)\sim(1,1-y)
$$
What would be a good way to approach this question? I have not had any success constructing a map between spaces

Edit: I remember that homeomorphism must preserve orientability. So this could be used to disprove a homeomorphism. 
The mobius band is non-orientable, as is the klein bottle. What I am not sure about is if we take the union of two non-orientable Mobius bands and attach their boundary circles, do we still get a non-orientable surface.
I think the gluing the boundary step may switch the orientabilty. So we have an orientable surface which therefore cannot be homeomorphic to the non-orientable Klein bottle.
I am unsure how to prove this in a formal way (with equations and notation etc)
Would appreciate your help

This is a diagram of the Klein bottle, note that the diagonal lines divide it into 2 Möbius strips sharing a boundary:

So the answer is yes.
