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I'm trying to make sure that I have correctly proved Munkres' Lemma 2.1, which is left to the reader. The lemma states:

Let $f: A \to B$. If there are functions $g: B \to A$ and $h: B \to A$ such that $g(f(a)) = a$ for every $a$ in $A$ and $f(h(b)) = b$ for every $b$ in $B$, then $f$ is bijective and $g = h = f^{-1}$.

Here is my attempted proof.

We will show that $f$ is bijective by showing that it is both surjective and injective. Fix $b \in B$. Then $h(b) = a$ for some $a \in A$. Applying $f$, we obtain $f(a) = f(h(b)) = b$, so $f$ is surjective. Now, suppose $f(a) = f(a')$ for some $a,a' \in A$. Applying $g$, we obtain $g(f(a)) = g(f(a'))$ and hence $a = a'$, so $f$ is injective and hence bijective.

Now, it suffices to demonstrate that $g = h$. We have $$g \circ \mathrm{id}_B = g \circ (f \circ h) = (g \circ f) \circ h = \mathrm{id}_A \circ h = h.$$ Therefore, we have $f \circ g = f \circ h = \mathrm{id}_B$ and $g \circ f = h \circ f = \mathrm{id}_B$, so $g = h = f^{-1}$, as required.

The thing I'm most concerned about is whether I've shown that $g = h = f^{-1}$. This requires showing two things, I believe: (1) that $g$ and $h$ are equal and (2) that they both operate as both left and right inverses. By showing that $g$ is also a right inverse, I've shown it is a two-sided inverse, and similarly by showing that $h$ is also a left inverse. I haven't shown that any two-sided inverse is necessarily equal to both $g$ and $h$, so the equality to $f^{-1}$ is not immediately clear to me, which suggests I should also prove (3) any other inverse is necessarily equal to $g$ and $h$.

Nico
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Cardinality
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    Suppose $g$ and $g'$ are two-sided inverses for $f:A\to B$ (so implicitly $f$ is bijective). For any $b\in B$, choose $a\in A$ so that $f(a) = b$. Then $a = g(f(a)) = g(b)$ and $a = g'(f(a)) = g'(b)$. Since $b\in B$ was arbitrary, $g = g'$. Incidentally, this only uses the fact that $g$ is a left inverse. – Nico Aug 08 '22 at 19:14
  • @Nico I understand this proof, thank you. My understanding from your comment is that it is necessary to prove this third additional fact? – Cardinality Aug 08 '22 at 19:17
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    Technically, it is necessary (as you note in your last paragraph). Though depending on your background you may eventually assume basic facts like these without proof. For now it may be useful to prove these basic claims until they eventually become second nature. My rule of thumb (which I don't always follow) is if I can't immediately see why it's obvious, I will try and prove whatever facts I leave implicit to see what it would take to spell out all the details (even if I don't write all those details out). Good on you for your thoroughness! – Nico Aug 08 '22 at 19:27
  • @Nico Thank you, I greatly appreciate it. If I could ask one more question: I was thinking that maybe it just followed immediately from $g = h$. Here's my idea. Let $\alpha, \beta: B \to A$ be two two-sided inverses for $f$. Then $\alpha$ is a left inverse and $\beta$ a right inverse. Taking $\alpha$ in place of $g$ and $\beta$ in place of $h$, $\alpha = \beta$, and the proof is complete. Does that work? – Cardinality Aug 08 '22 at 19:48
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    That's right, and more streamlined than the other argument! – Nico Aug 08 '22 at 19:51
  • That's great. I think it'd seem clear if you add $(i)\ g(f(a))=a$ implies $g$ is the left inverse of $f$ and $(ii)\ f(h(b))=b$ implies $h$ is the right inverse of $f$ before the line $$g=g \circ \mathrm{id}_B = g \circ (f \circ h) = (g \circ f) \circ h = \mathrm{id}_A \circ h = h.$$ – Myo Nyunt Sep 04 '22 at 13:22

2 Answers2

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The "function notation" $f^{-1} : B \to A$ is only used for bijections $f : A \to B$. But what is the precise definition of $f^{-1}$?

For $b \in B$ one defines $f^{-1}(b)$ as the unique element $a \in A$ such that $f(a) = b$. Note that the existence of such $a$ follows from the surjectivity of $f$ and the uniqueness of $a$ from the injectivity of $f$.

By definition of $f^{-1}$ one has $f^{-1}(f(a)) = a$ for all $a \in A$ and $f(f^{-1}(b)) = b$ for all $b \in B$, i.e. $$f^{-1} \circ f = id_A \text{ and } f \circ f^{-1} = id_B \tag{*}.$$

You proved that $g \circ f = id_A$ and $f \circ h = id_B$ imply that $f$ is a bijection. Applying this (mutatis mutandis) to $(*)$, we see that $f^{-1}$ is a bijection. Of course you can also derive this directly from the definition of $f^{-1}$. That $f^{-1}$ is a bijection may seem to be obvious, but actually it requires a (simple) proof.

You also proved $g = h$, but $g = h = f^{-1}$ in fact requires a proof. So let us do it in one go.

$$g = g \circ id_B = g \circ (f \circ f^{-1}) = (g \circ f) \circ f^{-1} = id_A \circ f^{-1} = f^{-1} \\ h = id_A \circ h = (f^{-1} \circ f) \circ h = f^{-1} \circ (f \circ h) = f^{-1} \circ id_B = f^{-1}. $$

Let us finally observe that the follwing are equivalent:

(1) $f$ is a bijection.

(2) $f$ has a two-sided inverse $\phi : B \to A$ (i.e. $\phi \circ f = id_A$ and $f \circ \phi = id_B$).

(3) $f$ has a left inverse $g$ and a right inverse $h$.

(1) $\implies$ (2) is obvious (take $\phi = f^{-1})$, as is (2) $\implies$ (3) (take $g = h = \phi)$. (3) $\implies$ (1) has been proved by you. You also proved (3) $\implies$ (2) by showing that $g = h$.

The main ingredient in (1) $\implies$ (2) is our above explict description of $f^{-1}$ which led to (*).

Also note that if a function $\phi$ as in (2) exists, then it is uniquely determined. In fact, if $\psi$ is any two-sided inverse, then $\psi = id_A \circ \psi = (\phi \circ f) \circ \psi = \phi \circ (f \circ \psi) = \phi \circ id_B = \phi$. As you have proved, both $f$ and $\phi$ are bijections, and they are inverse to each other. By the way, this also shows that our $f^{-1}$ is the only two-sided inverse for $f$.

Similarly, if $g$ and $h$ as in (3) exist, then they are uniquely determined and equal.

Paul Frost
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This exercise does not require anything more than understanding definitions.

If $b\in B$ then there is some $h(b)=a\in A$ such that $f(a)=b$, so $f$ is surjective. If $p,q\in A$ and $f(p)=f(q)$ then $g(f(p))=p=g(f(q))=q$ so $p=q$, thus $f$ is injective.

Given $a\in A$, $f(a)=b\in B$ then $g(f(a))=a$ so $g(b)=a$, and $f(h(b))=f(a)$, hence from bijectivity $h(b)=a$, hence $g\equiv h\equiv f^{-1}$.

Suzu Hirose
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