 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2007/4/c/solution

Because the image is integer-valued, the open interval

(n - \frac{1}{2}, n + \frac{1}{2})

contains exactly one possible image value, namely

n

. Thus

f^{- 1} ({n}) = f^{- 1} ((n - \frac{1}{2}, n + \frac{1}{2})) .

The right side is open in

A

f^{- 1} ({n}) is open in A .

This includes the possibility of an empty fiber. The source notation

f^{- 1} (n)

is understood as the preimage of the singleton

{n}

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