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.

Proof

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• Note to self - uniqueness comes from that we're factoring through a surjective map (namely, $\pi$), we only really have to show the result is injective.

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