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Title says it all.

To give a concrete example:

Let $X$ and $Y$ be non-empty sets and $f: X \rightarrow Y$ a mapping. Prove that for relation defined by $\{(x_1,x_2) \in X^2 : f(x_1) =f(x_2)\}$ reflexivity holds.

I'm not sure if it is enough to simply state that by definition every $x$ has only one image $f(x) \rightarrow f(x)=f(x)$, which is our original definition.

Appart from probably different eqvivalence classes, does it make any difference when the mapping is f.i. injective or constant ($\exists c \in Y (\forall x \in X: f(x)=c)$)?

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up vote 2 down vote accepted

$\forall x \in X, \;f(x) =f(x)$, so reflexivity holds, i.e., $(x, x) \in \mathcal{R}.$

If $f(x) = c$, for some constant c, then $f(x) = f(x) = c$.

Note that for all properties of an equivalence relation, it doesn't matter if $f$ is injective, or $f(x)=c$ for some constant $c$. You should find that the relation you defined IS an equivalence relation:
$\mathcal{R}\;$ is

  • reflexive,
  • symmetric, and
  • transitive.

It's more an exercise in using the definition of an equivalence relation and what is required to meet each of the properties of an equivalence relation, and less about the function $f$.

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I think you understood this; it is trivially reflexive, and I suspect that you may have simply second-guessed yourself because it struck you as obvious? – amWhy Nov 26 '12 at 23:41
Yes, genau this was the problem... But such examples are best to test your understanding. In the process I got confused and thought that my proof depends on type of the mapping even though I could see that the relation must be reflexive (and yes, apart from that also symetric and transitive... but the two proofs made me no difficulty). And just to make sure: for f(x)=c there is just one eqvivalence class whereas for injective f, there are as many eqvivalence classes as there are such values for y (that corespond to different values of x). – smihael Nov 27 '12 at 0:03
$\ddot\smile$ for my slept friend. – Babak S. Sep 1 '13 at 11:11

A relation $R$ is reflexive if and only if $xRx$ for all $x \in X$. Since $f(x) = f(x)$ for all $x \in X$ the relation is reflexive. Injectivity or constancy do not affect things.

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