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I'm having trouble proving the following:

1) Prove that a function f is one-one if and only if ∀xy∈A,x≠y→f(x)≠f(y).

I was told to do this by the contrapositive, so I show that f(x)=f(y) → x=y. However, how do I do this? Can I make an example?

2) Assume f:A→B is a correspondence (or bijection), i.e., f is one-one and onto. Show that f has a unique inverse (which we'll denote by f^(−1)), and that ∀x∈B,f(f^(−1)(x))=x, i.e., that f is f^(−1)'s inverse.

I'm unsure how to approach this one.

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    What is your definition of an injective function? – 2017-02-05
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    For the first problem, an example is not a proof. You need to show that the given statement is the same as the contrapositive of the definition. – 2017-02-05
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    The first question is actually the _definition_ of a function being one-to-one. The contrapositive is a very common way of showing that a specific, given function is one-to-one, i.e, assume that two function values are equal and show that they must have come from the same point in the domain, conclude that the function is one-to-one. So I don't understand the first question as you've posed it. – 2017-02-05
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    1) is a direct equivalence of the definition... 2) can be approached by first noting that a pseudo-inverse of an injective function is surjective, and vice versa. To show uniqueness you can rely on associativity of function composition (Suppose g, h are both inverses, then so on and so forth...) – 2017-02-05
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    For the second question, can you define a function $g: B \to A$ in such a way that $g$ is the inverse of $f$? You'll need to show that $g(f(x)) = x$ for all $x \in A$ and $f(g(x)) = x$ for all $x \in B$ in order to conclude that the $g$ you produced is indeed $f^{-1}$. – 2017-02-05

2 Answers 2

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I suppose your original definition is (for some $f:X \rightarrow Y$)

$$f \text{ injective iff } \forall x,y \in X: (f(x) = f(y))\rightarrow (x= y)\text{.}$$

Then the other statement is just the contrapositive, indeed,as $p \rightarrow q$ is logically equivalent to $\lnot q \rightarrow \lnot p$. If you had logic before ,you can just stop there.

Otherwise a simple proof by contradiction will do: let $x \neq y$ be points of the domain $X$. Suppose that $f(x) = f(y)$. Injectiveness (the original definition) now allows us to conclude that $x=y$, but we cannot have both $x=y$ and $x \neq y$, contradiction, hence the assumption $f(x) = f(y)$ was wrong hence $f(x) \neq f(y)$.

On the other hand, if $f$ satisfies the $\neq$-variant then $f$ is injective using the same ideas: suppose $x,y$ satisfy $f(x) = f(y)$. Assume $x \neq y$. Then we could conclude $f(x) \neq f(y)$, a direct contradiction with $f(x) = f(y)$, so that assumption is wrong and so $x = y$, as required.

When $f: A \rightarrow B$ is a bijection ,for every $b \in B$ define $f^{-1}(b)$ to be the unique point $a \in A$ such that $f(a) = b$. There is at least one, by surjectiveness and at most one by injectiveness (if $a'$ is another candidate, they both map to $b$ so are the same). Now check whatever your definition of inverse of $f$ is (you don't state it, though I can guess).

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For the first part, prove the theorem directly from the definition of an injective function: a function is injective if (and only if, though the latter phrase is often left out in definitions) $f(x)=f(y)$ implies $x=y$.

But $f(x)=f(y)$ implies $x=y$ if and only if $x \not = y$ implies $f(x) \not = f(y).$ This is the application of contrapositive to the definition of one-to-one (injective) functions.

For the second part, you must first show that $f(x)$ has an inverse, then show that this inverse is unique. Since $f(x)$ is injective onto its range, $f(x)=f(y)$ implies $x=y.$

Designate a function $f^{-1}$ on the range element $f(x)$ in such a way that $f^{-1}(f(x))=x.$ We must show the function is well-defined, that is, there is no more than just $x$ as the answer. But $f(x)=f(y)$ if and only if $x=y,$ so $f^{-1}(f(x))=f^{-1}(f(y))=x.$ Hence $f^{-1}$ is well-defined. Similarly, $f^{-1}(f(x))=f^{-1}f(y))$ implies $x=y,$ from the definition of inverse, but from the one-to-one property of $f,$ we have $f(x)=f(y)$ so that $f^{-1}$ is itself also a one-to-one function. Finally, $f(x)$ maps onto all the range, so that $f^{-1}(x)$ maps onto all the domain, showing $f^{-1}$ is a bijection.

Now suppose $f(f^{-1})(x)=y.$ Then applying $f^{-1}$ to both sides gets $f^{-1}(y)=f^{-1}(x)$ which means $y=x$ since $f^{-1}$ is one-to-one. Hence $f(f^{-1})(x)=x.$

For uniqueness, suppose $f^{-1}(f(x))=x=g(f(x))$ for all $x.$ Then $f^{-1}(f(x))-g(f(x))=0$ for all $x,$ meaning $f^{-1}$ and $g$ are the same function. So the inverse must be unique.