Factoring a function through the projection of an equivalence relation induced by that function yields an injection

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Statement

$\xymatrix{ X \ar[r]^f \ar[d]_{\pi} & Y \\ \frac{X}{\sim} \ar@{.>}[ur]_{\tilde{f} } }$
Commutative diagram showing the situation

Let $X$ and $Y$ be sets, let $f:X\rightarrow Y$ be any function between them, and let $\sim\subseteq X\times X$ denote the equivalence relation induced by the function $f$, recall that means:

• $\forall x,x'\in X[x\sim x'\iff f(x)=f(x')]$

Then we claim we can factor $f:X\rightarrow Y$ through $\pi:X\rightarrow \frac{X}{\sim}$ to yield a unique injective map:

• $\tilde{f}:\frac{X}{\sim}\rightarrow Y$

Furthermore, if $f:X\rightarrow Y$ is surjective then $\tilde{f}:\frac{X}{\sim}\rightarrow Y$ is not only injective but surjective to, that is: $\tilde{f}:\frac{X}{\sim}\rightarrow Y$ is a bijection.

Applications

Topology:

• If $f:X\rightarrow Y$ is a continuous map then factoring it through the projection of the equivalence relation it induces yields a continuous injection - topological version of this theorem almost exactly. We can extend this slightly in the case $f:X\rightarrow Y$ is surjective:
  • If $f:X\rightarrow Y$ is continuous and surjective then factoring it through the canonical projection of the equivalence relation it induces yields a continuous bijection - this is an extension of the previous statement, recall that when we take a surjection and apply passing-to-the-quotient we get surjection, we know from above it's injective. So now it's bijective

Proof

Outline of proof method

1. We must check the set up satisfies the requirements of the passing-to-the-quotient theorem (to yield $\tilde{f}:\frac{X}{\sim}\rightarrow Y$).
2. We apply the theorem to get $\tilde{f}:\frac{X}{\sim}\rightarrow Y$.
3. Next we must show $\tilde{f}:\frac{X}{\sim}\rightarrow Y$ is injective

Proof body

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Part 3 was the only non-trivial part. It is obvious that we can factor $f$ through $\pi$, in fact:

• To do so requires that "$f$ be constant on the fibres of $\pi$", ie:
  • $\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)]$ (see Equivalent conditions to being constant on the fibres of a map)
    • But if $\pi(x)=\pi(y)$ then $x\sim y$ and by definition $x\sim y\iff f(x)=f(y)$! Trivial!

Oh just FYI there's a commented out note and a requires-proof template at the top of the ==proof== heading, remove that later!

Step 3

We wish to show that $\tilde{f}:\frac{X}{\sim}\rightarrow Y$ is injective, ie: $\forall a,b\in\frac{X}{\sim}[\tilde{f}(a)=\tilde{f}(b)\implies a=b]$

• Let $a,b\in\frac{X}{\sim}$ be given.
  • Suppose $\tilde{f}(a)\ne\tilde{f}(b)$ then by the nature of logical implication we do not care whether or not $a=b$, either way it is true, we're done in this case.
  • Suppose $\tilde{f}(a)=\tilde{f}(b)$, by the nature of logical implication we now require this leads to $a=b$
    • By the surjectivity of $\pi:X\rightarrow\frac{X}{\sim}$ we see:
      • $\exists x\in X[\pi(x)=a]$ and $\exists y\in X[\pi(y)=b]$
        • Note that $\tilde{f}(a)=\tilde{f}(\pi(x))$ and $\tilde{f}(b)=\tilde{f}(\pi(y))$, also by assumption we have $\tilde{f}(a)=\tilde{f}(\pi(x))= \tilde{f}(\pi(y))=\tilde{f}(b)$
        • Notice also that $f(x)=\tilde{f}(\pi(x))$ and $f(y)=\tilde{f}(\pi(y))$ (this is the whole point of $\tilde{f}$)
          • So we see $f(x)=\tilde{f}(a)=\tilde{f}(\pi(x))= \tilde{f}(\pi(y))=\tilde{f}(b)=f(y)$, or importantly:
            • $f(x)=f(y)$
          • Recall that $x\sim y\iff f(x)=f(y)$, so now we know $x\sim y$
        • An elementary property of $\pi:X\rightarrow\frac{X}{\sim}$ is that if $x\sim y$ then $\pi(x)=\pi(y)$
          • Explicitly, notice we have: $\pi(x)=\pi(y)$
        • Recall that $a=\pi(x)$ and $b=\pi(y)$ by definition of $x$ and $y$, so:
          • $a=\pi(x)=\pi(y)=b$ or more simply: $a=b$
    • We have shown that supposing $f(a)=f(b)$ is given then $a=b$
  • Both cases have been evaluated and the implication holds for both
• Since $a,b\in\frac{X}{\sim}$ were arbitrary we have shown this for all $a,b\in\frac{X}{\sim}$

See also

• If a surjective function is factored through the canonical projection of the equivalence relation induced by that function then the yielded function is a bijection
• Factoring a continuous map through the projection of an equivalence relation induced by that map yields an injective continuous map
  • If a surjective continuous map is factored through the canonical projection of the equivalence relation induced by that map then the yielded map is a continuous bijection