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OhMyMarkov
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The title says it: why is the following not true: $f(f^{-1}(B))=B$?
Thanks!
Thanks!
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You have not given a context for this, but in general if you have a function $f:A\to B$ there is no reason to suppose that the range of $f$ is the whole of $B$. If $f$ is not surjective then $f(f^{-1}(B)) = f(A) \ne B$.OhMyMarkov said:The title says it: why is the following not true: $f(f^{-1}(B))=B$?
Thanks!
This statement is not always true because the inverse of a function does not always exist. When the inverse function does exist, it may not necessarily map all elements from the range of f back to their original values in the domain. Therefore, f(f^(-1)(B)) may not always equal B.
Yes, it is possible for f(f^(-1)(B)) to equal B in certain cases. If the function f is bijective, meaning it is both injective (one-to-one) and surjective (onto), then the inverse function exists and f(f^(-1)(B)) will equal B.
To determine if a function is bijective, you can check if it passes the horizontal line test. This means that no horizontal line intersects the graph of the function more than once. If this is true, then the function is both one-to-one and onto, and therefore bijective.
Yes, there are other cases where f(f^(-1)(B)) can equal B. If the function f is an identity function, meaning it maps every element in its domain to itself, then f(f^(-1)(B)) will equal B. Additionally, for some functions, the domain and range may be the same set, in which case f(f^(-1)(B)) will also equal B.
Finding the inverse of a function is useful for solving equations, finding the inverse of a matrix, and for certain applications in mathematics and physics. It also helps in understanding the relationship between the domain and range of a function.