Consider a [[function]] $f : X \to Y$. Then we can construct the [[equivalence]] relation $ x \sim x' \iff f(x) = f(x') $ ie points that map to the same output. This inherits the desired properties from the "equals" relation. i.e. the equivalence classes are the [[preimage]]s $f^{\text{pre}}(y)$ for each $y$ # group structure Consider $g$ an [[automorphism]] of $X$ such that $f(g(x)) = f(x)$ for all $x$. Then we say $f$ is **invariant** to $g$. The set of such $g$ forms a subgroup $\mathrm{Ivnt}(f) \subseteq \mathrm{Aut}(X)$ (under composition): - $g, g' \in \mathrm{Ivnt}(f) \implies g \circ g' \in \mathrm{Ivnt}(f)$ - $\mathrm{id}_{X} \in \mathrm{Ivnt}(f)$ - $g \in \mathrm{Ivnt}(f) \implies [\forall x: f(x) = f(g(g^{-1}(x))) = f(g^{-1}(x))] \implies g^{-1} \in \mathrm{Ivnt}(f)$ This also gives us another equivalence relation on $X$: ie points that can be transformed into each other under some invariant: $ x \sim_{\mathrm{Ivnt}(f)} x' \iff \exists g \in \mathrm{Ivnt}(f) : g(x) = x' $ This is clearly [[reflexive]], symmetric, and transitive. Note that $x \sim_{\mathrm{Ivnt}(f)} x' \implies x \sim_{f} x'$. --- Now suppose $X = \mathbb{R}^{D}$ is a [[vector space]] and so we can [[algebraic representation|represent]] each $g$ as an [[invertible linear map|invertible]] $D \times D$ matrix. eg [[Lie group]]s for [[spatial invariance]] # sources [[1995VanGoolEtAlVisionLiesApproach|Vision and Lie's approach to invariance]] [[2014CohenWellingLearningIrreducibleRepresentations|Learning the Irreducible Representations of Commutative Lie Groups]]