Proof strategy for $(\Leftarrow)$: If $g \circ f = id_A$, then $f$ onto $\Leftrightarrow$ $g$ 1-1. [Chartrand 3Ed P239 9.72]

Question

For nonempty sets $A$ and $B$ and functions $f \colon A \to B$ and $g \colon B \to A$, suppose that $g \circ f =$ the identity function on $A$. $(♦)$

(e) $(\Leftarrow)$ Assume that $g$ is one-to-one. Because $g$ is a function, for all $b \in B$, there exists $a \in A$ such that $a = g(b)$. Apply $\color{orangered}{f}$ to both sides: $\color{orangered}{f}(a)= \color{orangered}{f}(g(b))$. Next, apply $g$ to both sides: \begin{align} g(\color{orangered}{f}(a))& =g(\color{orangered}{f}(g(b))) \\ \text{ thanks to } (♦) &= g(b). \end{align} $g$ is given as one-to-one, so $f(a)=b$ and so $f$ is onto.

What's the proof strategy? This isn't a duplicate of https://math.stackexchange.com/a/750495/53259. I'm posting de novo because this direction looks wilier and more guileful. I'm not asking about the proof or formal arguments. For example, how would one determine when to apply $f$ or $g$?
I realise that the proof leverages $(♦)$. Are there pictures?

Answer 1

The overall proof strategy is called "chasing definitions". For the first three steps there is only one thing to do, so we follow the strategy of "fitting square pegs in square holes".

So let's begin:

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Definition of $f:A\rightarrow B$ onto is that for every $b\in B$ there exists $a\in A$ such that $f(a)=b$.

So we have to start with $b\in B$ and then find a suitable $a$. So

Let $b\in B$.

What can we do with a $b\in B$? Look at what we have: we have $b\in B$, we have a function that takes things in $A$ to $B$, and a function which takes things in $B$ to $A$. We have one square peg, and one square hole. The only thing we can do is apply $g$. So we apply $g$.

Consider $g(b)$. Now $g: B\rightarrow A$ so $g(b)=a$ for some $a\in A$.

Okay now we have an $a\in A$. What can we do with an $a\in A$? We have one round peg and one round hole. The only thing we can do is apply $f$.

Applying $f$, we obtain $f(g(b))=f(a)$

Now we have a bunch of $f$'s on the outside. If we knew that $f\circ g$ were identity we would be done (remember our goal? we are trying to get $a=f(b)$). But we only know that $g\circ f = \mbox{id}_A$. To apply that tool we need a $g$ on the outside. Our other tool is that we are allowed to cancel $g$'s on either side, because $g$ is $1-1$. Again, this requires $g$ on the outside. To apply either of our tools then, we need $g$ on the outside. So we apply $g$.

Applying $g$, we obtain $g(f(g(b)))=g(f(a))$

Alright, now we are in a situation where both of our two tools apply. So we have to choose which tool to use. Obviously we shouldn't use the tool that cancels $g$ on both sides, because that takes us back to the previous step (don't walk backwards). So we should use the tool that we can cancel $g\circ f$. But where should we use it? If we cancel out the right side then we've run out of outside $g$'s and we won't be able to use our other tool (don't go down a dead-end street). We want to use our tool but still have a $g$ left on each side to apply our other tool on. So we cancel out $g$ on the left side.

Applying ♦ we have $g(b)=g(f(a))$

Now we have a $g$ on both sides and can apply our 1-1 tool.

So since $g$ is 1-1 we have $b=f(a)$.

Which was our goal, so we stop.

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Notice that once we get to the fourth step, we have to start deciding which hypotheses we should apply: we have two of them, one is that $g\circ f$ is identity and the other is that we can cancel out $g$'s when they appear on both sides. When we have a choice between two tools, a way to choose one is to choose not the other. In this case it was clear which tool we shouldn't use, which left us with only one other tool to apply. To make our decisions we applied two maxims: don't go down a dead end street and don't walk backwards.

Answer 2

You want to show that from $$ g\circ f=\mathit{id}_A \qquad\text{and}\qquad g \text{ one-to-one} $$ you can deduce that $f$ is onto.

The first obvious thing to do is considering an element $b\in B$: we need some $a\in A$ such that $f(a)=b$. So the second step is trying with $g(b)\in A$: we can do with what we have available, in this case $f$, $g$ and $b$. So the only candidate for $a\in A$ with $f(a)=b$ is $a=g(b)$.

Let's apply the hypothesis to $a$: $a=g(f(a))$, which can be rewritten as $$ g(b)=g(f(a)). $$ Since $g$ is one-to-one, $b=f(a)$. QED

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About the strategy, there's not very much to say: since you want to prove that $f$ is onto, you must start from an element $b\in B$, don't you? What can you do with $b$? You can just consider $g(b)$, for there's nothing else available. Then if we are to find $a\in A$ such that $b=f(a)$, the only candidate is $a=g(b)$.

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If you have already proved that a one-to-one map has a left inverse, the proof can be carried on algebraically. Call $h\colon A\to B$ a left inverse of $g$, that is, $h\circ g=\mathit{id}_{B}$. Then $$ f=\mathit{id}_B\circ f=(h\circ g)\circ f=h\circ(g\circ f)= h\circ\mathit{id}_A=h $$ But a left inverse is always onto.

Answer 3

So we know $g \circ f = 1_A$. We want to show $f$ is onto iff $g$ is 1-1.

There is not much one can do: show two implications.

So suppose $f$ is onto. We want to show $g$ is 1-1. This one was already handled in this answer.

So suppose $g$ is 1-1, we want to show $f$ is onto. So start with a $b \in B$, we want to find some point in $A$ that maps to $b$ under $f$. We want to use $g$, which goes from $B$ to $A$, so consider $a = g(b)$, we hope that $f(g(b)) = b$. We haven't used that $g \circ f = 1_A$ yet, so let's consider $g(f(g(b))) = g(b) = a$. But look, we have two points, $f(g(b))$ and $b$, that under $g$ both map to the same value $a$. By 1-1-ness $f(g(b)) = b$, so our hopes are vindicated: $g(b)$ is mapped under $f$ to our target $b$ and so $f$ is onto.

Note that in every step I just looked for one of the givens (the identity or the 1-1-ness) to be usable. The hope that $g(b)$ would be the correct point, is based partly on "well, what else can we do" and the intuition that $g$ and $f$ already are "half-inverses" already. Adding the extra assumptions on one of them actually makes the real, two-sided, inverses.

Answer 4

The strategy should be to prove both directions with a minimum number of motions. (It seems that $\Rightarrow$ has been proven in an earlier question.)

We are given $f: \ A\to B$, $\>g:\ B\to A$, such that $\ g\circ f={\rm id}_A$.

Claim: $\ f$ is onto $\ \Leftrightarrow\ g$ is one-one.

Proof of $\Rightarrow$: $\ $ Assume $f$ is onto and $a=g(b)$ for some $b\in B$. By assumption on $f$ there exists $a'$ with $b=f(a')$, which implies $a=g(b)=g\circ f(a')=a'$. It follows that necessarily $b=f(a)$, whence $b$ is uniquely determined by $a$.

Proof of $\Leftarrow$: $\ $ Assume $g$ is one-one, and let a $b\in B$ be given. Put $g(b)=:a$ and $f(a)=:b'$. It follows that $g(b')=g\circ f(a)=a$. Since $g$ is one-one we conclude that $b=b'=f(a)$. This proves that $f$ is onto.

Answer 5

The strategy of the proof is to apply the definition of an injection and a surjection. Now one want to prove that $$ g\ \text{is an injection}\ \Rightarrow\ f\ \text{is a surjection}, $$ and given that $$ g\circ f=1_A. $$ It seems that here you have two points: '$g\circ f=1_A$' or '$g\ \text{is an injection}$'. In addition, one's target is to prove $f$ satisfy the condition in the definition '$\forall y_0\in B,\ \exists \epsilon\in A$, such that $f(\epsilon)=y_0$'. So actually you have to start from this 'any $y \in B$' to your target: $\epsilon\in A$.

Then the strategy is clear: Apply the functions and connect the two conditions given. Before you apply the function, look at the which set the variable is in.

So here $y_0\in B$, then apply, we get $g(y_0)$. We can't do anything with this single expression. Then continue to apply functions.

Notice that now $g(y_0)\in A$, so apply $f$. This gives $f(g(y_0))$. We still can't do anything about it. Then apply functions again.

The codomain of $f$ is $B$, so $f(g(y_0))\in B$. Then we have to apply $g$ again, giving $$ g(f(g(y_0))). $$ Look at this expression. It seems a bit messy, but now it's in a form of '$g(f(*))$', we know we can use one of the given conditions now: $$ g(f(g(y_0)))=(g\circ f)(g(y_0))=1_A(g(y_0))=g(y_0). $$ Now after we apply the first given condition, we notice that the two sides of the equation above have something in common: '$g(*)=g(\$)$'. Now we can use the other condition given, i.e., $$ f(g(y_0))=y_0. $$ Now it's clear: for any $y_0\in B$, we have $\epsilon=g(y_0)\in A$, such that $y_0=f(\epsilon)$. So we have found the '$\epsilon\in A$' and we can say that $f$ is a surjection.

I hope this helps.