Let $f$ be a function from a set $X$ into a set $Y$.


i) the function $f$ has an inverse if and only if $f$ is bijective ii) let $g_1$ and $g_2$ be functions from Y into X. If $g_1$ and $g_2$ are both inverse function of f, then $g_1 = g_2$, that is $g_1(y) = g_2(y)$ for all $y\in Y$

attempt at i)

--> if $f$ has an inverse this means there exists a function g of $Y$ s.t. $g(y) = x\in X$ but this just means that for every $y\in Y$ there exists an $x\in X$ s.t $f(x) = y$ therefore the function is surjective.

to show that the function is injective:

same assumption of f having an inverse.

Assume $f(x_1) = f(x_2)$ these are elements of Y. taking the inverse function g of both of them: $$\implies \\g(f(x_1)) = g(f(x_2))$$ $$\implies \\x_1 = x_2$$ which shows the function is injective.

Therefore the function is bijective.

Now to go the other way and show the function f has an inverse:

Assuming the function is bijective. By the property of surjectivity, for each $y\in Y$ there exits a $x\in X$ s.t $f(x) = y$ Does this mean that there must exist a function g(i.e the inverse) that would allow each $g(f(x)) = x$? if so that would be the conclusion.

Attempt at ii)

well if $g_1(y)$ and $g_2(y)$ are both inverse functions of f, then take the difference of the two functions:

$$f(x) - f(x) = g_1(y) - g_2(y)$$ $$ 0 = g_1(y) - g_2(y) $$ $$ g_2(y) = g_1(y)$$

now I think I may have the right idea for this one, but I know my notation has to be off because there is no way that $f(x) = g_1(y)$, how would I express this?

  • $\begingroup$ While this has implications in topology, I wouldn't say that this is a topological question (for example, it also has implications in group theory, combinatorics, real analysis, or pretty much anything). $\endgroup$ – GPerez May 1 '15 at 15:34
  • $\begingroup$ what would be a better heading to put it under @GPerez ? $\endgroup$ – dc3rd May 1 '15 at 15:37
  • $\begingroup$ As functions between sets without any additional structure, I might consider these set-theoretic properties of functions. However set theory usually refers to axiomatic systems, logic, etc. Here I'd just say properties of general functions. $\endgroup$ – GPerez May 1 '15 at 15:48

For the end of i), You can construct the inverse explicitly: $g(y) = x$, where $x$ is such that $f(x) = y$. Now you need to use injectivity to show that this is well defined.

For ii), your approach cannot be right, since $Y$ is a set with no addition defined. A way to start the proof could be, for example $$g_1\circ f = g_2\circ f $$ Now use the fact that $f\circ g_1=\mathrm{id}_Y$ (compose on both sides with $g_1$).

  • $\begingroup$ for the first part of your answer, I guess to use injectivity and the fact that I also know that the function f exists, could I assume $g_1(y) = g_2(y)$ and then apply the function f to both sides? as for the second one, I'm not fully clear how I could do that, since I have to show $g_1 = g_2$ $\endgroup$ – dc3rd May 1 '15 at 15:54
  • $\begingroup$ @dc3rd I think you're mixing things up, the first part of the question has no $g_1,g_2$. You want to prove that a certain $g:Y\to X$ exists, and said $g$ is to be the inverse of $f$. I've given a possible definition of $g$, but the problem is that there could be more than one $x$ that maps to $y$. To show that this isn't the case you have to use injectivity. The second part is really just doing exactly as I said: compose with $g_1$ on both sides of the equality I've written. $\endgroup$ – GPerez May 1 '15 at 16:00
  • $\begingroup$ For the first we constructed our function, and since we have assumed injectivity then it would apply to our function? I'm having problems understanding how to apply it specifically. So for the second you mean : $g_1(g_1\circ f )= g_1(g_2\circ f)$ would this mean $g_1(y) = g_1(y)$. but that is only the $g_1$ function $\endgroup$ – dc3rd May 1 '15 at 16:14

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