# A generalized diagonal?

A simple question. Let $f:X\to Y$ be a function and let $E_f:=\{(x, y): f (x)=f (y)\}\subset X\times X$. What is the name of the set $E(f)$? It would be nice to have some reference also. It seems to be a well known notion/construction in set theory or topology, but I couldn't find anything about it.

Another question is whether the following statement is true: If $X$ is a connected topological space and the function $f$ is continuous then $E_f$ is connected.

• In algebra, if f is a homomorphism, then E is a congruence. If f is not a homomorphism, E can still be an equivalence relation of interest. – The Masked Avenger Dec 28 '14 at 19:51
• Another term for this is equalizer (ncatlab.org/nlab/show/equalizer) where the two parallel morphisms coincide. – Vidit Nanda Dec 28 '14 at 20:44
• @ViditNanda it's not the equalizer, it's the kernel pair. – user40276 Dec 30 '14 at 0:21

It's called the kernel or kernel pair of $f$. It is used all over the place in category theory, for example to describe the useful notion of regular category where one sets up Galois connections which in one direction sends a morphism $f: X \to Y$ to the pair of projections $\pi_1, \pi_2: \ker(f) \rightrightarrows X$, and in the other sends a parallel pair $K \rightrightarrows X$ to its coequalizer $X \to Q$. The image of $f: X \to Y$ in such categories is then the coequalizer of its kernel pair.

E(f) does not have to be connected even when $\ X\$ is.

Example: Consider $\ S^1 := \{z\in\mathbb C : |z| = 1\}\$ -- the unit circle; and also $\ f:S^1\rightarrow S^1\$ such that:

$$\forall_{z\in S^1}\ f(z):= z^2$$

Then $\ E(f) = \{(u\ v)\in S^1\times S^1 : u^2=v^2\}\$ is not conected.

REMARK:   If $\ f:X\rightarrow Y\$ is such that $\ X\$ is connected, and $\ f^{-1}(y)\$ is connected for every $\ y\in Y\$, then $\ E(f)\$ is connected

Example--just a variation of the above one:   E(f) is disjoint for $\ f:\mathbb R\rightarrow\mathbb R^2$ given by: $\ \forall_{x\in\mathbb R}\ f(x):= \exp(\imath\cdot x)$.

• A nice example! What is about $f:\mathbb{R}^n\to\mathbb{R}$. My original problem comes from a rather concrete $f:\mathbb{R}^4\to\mathbb{R}$ for which the statement is true, even in a considerably larger class of functions. – Vladimir Tkachev Dec 29 '14 at 20:06
• @VladimirTkachev -- you got nice result! (I tried for more of something positive but didn't get anything). – Włodzimierz Holsztyński Dec 30 '14 at 7:35
• The examples I have in mind, all of them have the property that "there exists at least one $a\in \mathbb { R }$ such that $f^{-1}(a)$ is connected". – Vladimir Tkachev Dec 30 '14 at 11:37
• P.S. Notice that the target space is $\mathbb {R}$, it might be crucial. – Vladimir Tkachev Dec 30 '14 at 11:44

$E_f$ is an equivalence relation on $X$ and conversely for every equivalence relation $R$, you can construct the topological quotient $X/R$ and for the induced map $f:X\to X/R$, $E_f=R$. There are unconnected equivalence relations on connected spaces that contradicts the above statement.

For example $X=[0,1], R=\{(x,x):x\in X\}\cup\{(0,1),(1,0)\}, X/R = S^1$.

I think this is called fibered product (or rather a special case thereof) $X\times_YX$. More generally, one would have two maps $X\to Z$, $Y\to Z$ and fibered product $X\times_ZY$.

@VladimirTkachev, in addition to the general case of maps $\ f:X\rightarrow Y,\$ was also concerned with the special case of $\ Y:=\mathbb R,\$ i.e. with $\ X\rightarrow\mathbb R.\$ I'll give an example (negative) in the extra simple case of $\ f:\mathbb I\rightarrow\mathbb R,\$ where $\ \mathbb I\$ is a closed interval.

EXAMPLE   Let $\ f:[-1;1]\rightarrow\mathbb R\$ be defined by:

$$\forall_{x\in[-1;1]}\quad f(x) := (x+1)\cdot x\cdot(x-1)$$

Then $\ E(f)\$ is disconnected. Indeed, $\ (-1\,\ 1)\in E(f)\$ is an isolated point of $\ E(f).\$ Of course $\ (1\,\ -\!1)\$ is another isolated point like this, and there are no other isolated points in $\ E(f)$.

PROOF   We have 3 cases:

• if $\ (x\ y)\in E(f)\$ is such that $\ f(x)=f(y) < 0\$ then both $\ x\ y>0\$ are positive;
• $\ E(f)\cap f^{-1}(0)\ =\ \{-1\,\ 0\,\ 1\}^2$
• if $\ (x\ y)\in E(f)\$ is such that $\ f(x)=f(y) > 0\$ then both $\ x\ y<0\$ are negative.

Thus the distance (say, Euclidean) from $\ (-1\,\ 1)\$ or $\ (1\,\ -\!1)\$ to any other point $\ (x\ y)\in E(f)\$ is at least $\ 1$.

End of PROOF

• Thank you for good examples! Probably it should be made more precise what kind of $X$ and $f$ are relevant. I've also considered a similar cubic polynomial, but with the domain of definition being the whole $X =\mathbb {R}$ instead of an interval. In that case, $E (f)$ is connected (being the union of an oval and a diagonal line passing through the oval). The example you construct can be thought as a cutoff of the set $E(f)$ above. Thus, some 'completteness' of X is needed. Another relevant example is $f=\sin x\sin y:\,\mathbb {R}^2\to\mathbb {R}$ for which $E(f)$ is connected again. – Vladimir Tkachev Jan 3 '15 at 13:59