Proving there exists a unique function decomposition

Question

I am trying to prove the following result.

Let $X$ and $Y$ be sets and $\sim$ an equivalence relation on $X$. Let $f: X \to Y$ be a function which preserves $\sim$, and let $\pi$ denote the projection $X \to X/\sim$. Prove that there exists a unique function $\tilde{f}$ such that $f = \tilde{f} \circ \pi$.

From the wording of the problem, it seems that I need to prove both existence and uniqueness, though I don't know how to prove uniqueness because that would require giving some sort of "formula" for $\tilde{f}$, but if I can give a formula for it, it must be unique. What the map is doing is: $$ x \stackrel{\pi}{\to} [x] \stackrel{\tilde{f}}{\to} Y. $$ Some versions of this decomposition I've seen send $[x]$ to an element of $\mathrm{Im}(f)$ and then inject into $Y$. I'm not sure if I need to include that because then I'd be adding an extra map. The $f$, as defined here, is only the composition of two maps.

As for proving uniqueness, it seems to me that this follows from the fact that a function is surjective if and only if it admits a right inverse. Suppose that $\tilde{f}$ and $\tilde{f}'$ have the property that $f = \tilde{f} \circ \pi = \tilde{f}' \circ \pi$. Cancelling on the right gives $\tilde{f} = \tilde{f}'$.

What am I missing?

Another thing I'm having trouble understanding is the exact meaning of a project map. The map sending $x \to [x]$ is surely a projection, but the definition I'd read is that $p$ is a projection if $p^2 = p$. I can't even compose $\pi$ with itself because $X \neq X/\sim$. Is there another, more general definition? It's it a shorthand for a surjective map?

Answer 1

If you define $\tilde f\colon X/\sim\longrightarrow Y$ by $\tilde f\bigl([x]\bigr)=f(x)$, does this make sense? That is, if $[x]=[y]$, do we have $\tilde f\bigl([x]\bigr)=\tilde f\bigl([y]\bigr)$? Yes, because$$[x]=[y]\implies f(x)=f(y)\iff\tilde f\bigl([x]\bigr)=\tilde f\bigl([y]\bigr).$$And then we have$$(\forall x\in X):\left(\tilde f\circ\pi\right)(x)=\tilde f\bigl([x]\bigr)=f(x),$$and therefore $\tilde f\circ\pi=f$.

And if $g\colon X/\sim\longrightarrow Y$ is such that $g\circ\pi=f$, then, if $x\in X$,$$g\bigl([x]\bigr)=\left(g\circ\pi\right)(x)=f(x),$$and so $g=\tilde f$.

Finally, the map $\pi\colon X\longrightarrow X/\sim$ is called the projection of a set onto the set of its equivalence classes with respect to the equivalence relation $\sim$. It has nothing to do with $p\circ p=p$, which is the concept of projection with respect to linear map from a vector space into itself.

Answer 2

I think a lot of your pain points can be remedied by recognizing that we often times identify equivalence classes by representatives. This makes $\pi : X \to (X/\sim) \simeq X$ at which point you can apply the projection definition you know (up to isomorphism).

By choosing a set of representatives (in the worst case by invoking the axiom of choice) we have (at least one) map $$g : X / \sim \to X$$ statisfying $\pi \circ g = id_{X / \sim}$ - precisely the right inverse you described. Using this we may define $\bar{f} : X / \sim \to Y, \bar{f} := f \circ g$ and we can show quite easily that this does statisfy the needed equality. Uniqueness is indeed trivial since $f=\bar{f}\circ \pi \iff \bar{f} = f \circ g$.