Suppose that $h : X → Y$ and $f : Z → Y$. Then $\exists g : X → Z$ s.t. $h = f ◦ g$ if $f$ is a bijection.

Question

Suppose that $h : X → Y$ and $f : Z → Y$. There exists a function $g : X → Z$ such that $h = f ◦ g$ if $f$ is a bijection.

I am not sure how to go about proving this. Please tell guide me.

proof: I understand the first step would go something like this and that essentially both sides are taking elements from $A$ and mapping them onto $C$.
Assume $f$ is bijective then it is both injective and surjective which means $f(b1)\neq f(b2)$ and that all elements of $C$ are mapped onto.

Answer 1

If f is bijective there is an inverse map $f^{-1}: Y \to Z $ with $f \circ f^{-1} = \text{Id}_Y$ and $ f^{-1} \circ f = \text{Id}_Z $. Define $g:= f^{-1} \circ h : X \to Z$.

Answer 2

You can easily prove it by constructing the desired application:

Since $f$ is bijection, then we can use $f^{-1}$ (if $f$ was not bijection, we could not do this). Then you can just rearrange the expression $h=f\circ g$ to obtain the explicit definition of $g$: $$h=f\circ g\implies f^{-1}\circ h=f^{-1}\circ f\circ g \implies \boxed{f^{-1}\circ h=g},$$ to be said, $g(x)=f^{-1}(h(x))$, wich is well defined due to $f$ being bijection.

Answer 3

Here a little trick to think of it in a different way.

For any $x_0 \in X$ we want $h(x_0) = f(g(x_0))$ and we are being asked to figure out what $g(x_0)$ is. (Well, actually you are just asked to prove that $g(x_0)$ exists... but we can do that be finding out what it is.) Now what if relabeled what $g(x_0)$ as $g(x_0) = y_0$?

[You might argue that $g(x)$ is a function and is a rule of mapping between values while $y$ is a single variable that has only one value. That's actually not a problem. We can (and frequently do) write functions as $y= g(x)$ where $y$ is a variable DEPENDENT upon $x$ [meaning for each value $x$ may take, $y$ will have a specific value]. But to avoid the issue and make this less abstract and vague I purposely chose $x_0$ to a fixed value so $g(x_0)$ (if it exists) will be a fixed value.]

So we have $h(x_0) = f(g(x_0)) = f(y_0)$ and we want to solve for what $y_0 = g(x_0)$ is. How do we do it?

The same as we always solve equations. We "unwrap" and isolate the $y_0$. So if $f(y_0) = h(x_0)$ how to we isolate and peel the $y_0$ out of its $f(y_0)$ wrapper? Well, we do the "undo" function of $f$. That is the inverse function $f^{-1}$.

It is essential that $f$ have and inverse $f^{-1}$ that is why $f$ must be injective; otherwise it would not.

So $f(y_0) = h(x_0)$ so $f^{-1}(f(y_0))=f^{-1}(h(x_0))$ so $y_0 = f^{-1}(h(x_0))$ and so $g(x_0)= y_0 = f^{-1}(h(x_0))$ and finally $g(x_0)= f^{-1}(h(x_0))$

Now if we can do such for any one $x_0\in X$ without relying upon any aspect about that particular $x_0$ we can do it for every $x\in X$. That is for every $x \in X$ we have $g(x) = f^{-1}(h(x_0))$ and so $g$ is function $g:x\to Y$ and $g = f^{-1}\circ h$

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Now if we want to make a more constructive proof using the definitions we can use that in the back of our mind. A proof could go like this:

$h:X\to Y$ is a function. So for every $x\in X$ there is a $y\in Y$ so that $y = h(x)$.

$f:Z\to Y$ is bijective function. As $f$ is surjective for every $y\in Y$ there are at least one $z_i\in Z$ so that $f(z_i) = y$. As $f$ is injective there is at most one so $z\in Z$ so that $f(z)=y$. And so for every $y\in Y$ there is exactly one $z\in Z$ so that $f(z) =y$. And as $f$ is a function every $z\in Z$ is such a value.

Well, as for every $x \in X$ there is a $y \in Y$ so that $y=h(x)\in Y$, and as for every $y \in Y$ there is distinct $z \in Z$ so that $f(z)=y$ then for every $x\in X$ there is a distinct $z \in Z$ so that $f(z) = h(x)$.

Thus we have described a well-defined mapping from $X\to Z$ where for each $x\in X$ there is a distinct $z\in Z$. In other words this is a function. Call it $f:X\to Z$.

Now $f\circ g:X\to Z\to Y$ if a composition function. $g(x)$ is the distinct $z\in Z$ so that $h(x)=f(z)$. So $f(g(z))= f(z) = h(x)$. So $h = f\circ g$. Done.