Relating with fix point theorem and continuity

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Homework Statement


Assume the function f : [0,1] x [0,1] -> [0,1] is continuous and apply the IVT to prove that there is a number c E [0,1] such that f(c,y0) = c for some y0 E [0,1]


The Attempt at a Solution


I tried to break the cube up with the ranging being y0 but I don't know how it maps y0 to [0,1] to be continuous, if I can prove this then I can use the IVT.

thankyou
 
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Let y0 be fixed. Consider the function g(x)=f(x,y0)-x. This function is sometimes \leq 0 and sometimes \geq 0. So you can apply the IVT on g.
 
After you apply the IVT on g, you will get that the number between is zero?
 
Yes. You will get that there exists a c such that g(c)=0. This will be the c you're looking for.
 
What do you mean when you mean y0 is fixed?
 
Just take y0 arbitrary.
 
So y0 in the element of [0,1] right? and one side I got g(x) >= 0 >= g(x)-1, Is this right?
 
Yes!
 
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