Quick question: Let $X$ and $Y$ bet two sets and $\sim$ an equivalence relation on $X$. I was wondering what it means to say that a function $f$: $X\to Y$ 'preserves' $\sim$ in this case. Does it mean that given two equivalent elements of $X$, their images under $f$ are identical? Thats what first comes to mind, but I just don't know. Thanks.

  • $\begingroup$ It would be best if you could come up with an example and context, but that would be my guess, as well. $\endgroup$ – Thomas Andrews Jan 25 '15 at 21:33
  • $\begingroup$ @ThomasAndrews I was looking for fun at a friend's notes[1] (I can't link directly to the file, so if you're interested, click on Math55a and go to Proposition 2.1 on page 6) on a math course he took last semester, and this general notion appeared, as stated, in a proposition about quotients. So there is not much else to 'contextualize' what I have more or less directly plagiarized. [1]: mit.edu/~evanchen/coursework.html $\endgroup$ – user210544 Jan 25 '15 at 21:49
  • $\begingroup$ @user210544 I just updated your my answer, in the given context it is what you guessed. $\endgroup$ – flawr Jan 25 '15 at 22:05

The preservation of equivalence relations under a function normally means

$$x \sim y \iff f(x) \sim f(y) \quad \forall x,y$$ but here $\sim$ is not defined on $Y$, so as far as I know there is no general notion of this, as long as e.g. $X\neq Y$ or $f$ isn't a homomorphism of a certain structure.

EDIT: In the given context I can confirm what you said, the author means $$x \sim y \iff f(x) =f(y).$$

You are considering the functions $f = \pi o \bar f: X \to Y$ where $\pi : X \to X/\sim$ and $\bar f : X/\sim \to Y$. This means that $\bar f$ (and therefore $f$ too) maps a whole equivalence class (the elements of $X/\sim$ are equivalence classes) to the same element in $Y$.

  • $\begingroup$ Try \sim instead of ~ to get $\sim$. $\endgroup$ – Cameron Williams Jan 25 '15 at 21:29
  • $\begingroup$ Well, you could think of sets equipped with equivalence relations, and the arrows between them are function preserving the relation (should be a category) $\endgroup$ – Marco Jan 25 '15 at 21:43

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