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What do we call a function whose domain and co-domain are the same set?

Edit: While i expressed my question in terms of functions, domains and codomains, i was actually interested in the most abstract mathematical formulation for similar structures. Thus i accepted as the correct answer endomorphism that, being at the level of category theory, answers rather the question: What do we call a morphism from a mathematical object to itself? I could find this out just after getting some answers

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    $\begingroup$ Transformation? That one has a geometric flavor. (I don't think that there is any standard name.) $\endgroup$ – Giuseppe Negro Jul 9 '15 at 21:50
  • $\begingroup$ Self-map. ${}{}$ $\endgroup$ – whacka Jul 9 '15 at 22:04
  • $\begingroup$ $\text{Endomorphism}$ $\endgroup$ – Alec Teal Jul 11 '15 at 12:42
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Endomorphisms: https://en.wikipedia.org/wiki/Endomorphism ${}{}{}{}{}{}{}$

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    $\begingroup$ I guess in $\textbf {Set}$ $\endgroup$ – Jorge Fernández Hidalgo Jul 9 '15 at 21:55
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    $\begingroup$ I think this works, although you may have to be explicit because in many contexts there is a notion of homomorphism and when you say endomorphism you may mean something stronger that a function. For example when saying $\varphi$ is an endomorphism on $S_3$ most people will not think $\varphi$ is just any function with domain and codomain $S_3$, but instead they will probably think $\varphi$ is a group homomorphism. $\endgroup$ – Jorge Fernández Hidalgo Jul 9 '15 at 22:01
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    $\begingroup$ Well, it depends on what kind of objects you're treating with, if it's just sets, the term is correct. If you're treating with groups, rings, etc. you'll have to be explicit about it. $\endgroup$ – YoTengoUnLCD Jul 9 '15 at 22:09
  • $\begingroup$ I like this answer because it is on the level of category theory, which was actually the level i had asked my question if i knew the correct abstract terms ... for example if i could use "morphism" instead of "function" and "mathematical object" instead of "set" $\endgroup$ – danza Jul 9 '15 at 22:11
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Whilst an endomorphism is a morphism or homomorphism from a mathematical object to itself, the technical term for a function that has a domain equal to it's co-domain is called an endofunction.

NB: A homomorphic endofunction is an endomorphism.

Edit: From Wikipedia:

Let $S$ be an arbitrary set. Among endofunctions on $S$ one finds permutations of $S$ and constant functions associating to each $x \in S$ a given $c \in S$.

Every permutation of $S$ has the codomain equal to its domain and is bijective and invertible. A constant function on $S$, if $S$ has more than $1$ element, has a codomain that is a proper subset of its domain, is not bijective (and non invertible). The function associating to each natural integer $n$ the floor of $n/2$ has its co-domain equal to its domain and is not invertible.

Finite endofunctions are equivalent to directed pseudoforests. For sets of size $n$ there are $n^n$ endofunctions on the set.

Particular bijective endofunctions are the involutions, i.e. the functions coinciding with their inverses.

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  • $\begingroup$ Oh, I like this idea, I don't know if it is commonplace but sounds interesting. $\endgroup$ – Jorge Fernández Hidalgo Jul 9 '15 at 22:03
  • $\begingroup$ This answer seems the most correct in mathematical terms, relatively to my question. Anyway, i really wanted to focus on the most abstract formulation, so i am tempted to accept endomorphism as the correct answer. Even though it is not technically the correct answer, it leads me better towards some useful theoretical framework, teaching me about a better formulation of my question $\endgroup$ – danza Jul 9 '15 at 22:14
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    $\begingroup$ Feel free to accept the other answer if you felt it helped you more. :-) $\endgroup$ – Zain Patel Jul 9 '15 at 22:15
  • $\begingroup$ I am also a bit confused by that text. As far as i understand an endofunction is always homomorphic, and it is always an endomorphism $\endgroup$ – danza Jul 9 '15 at 22:30
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    $\begingroup$ @danza: for example, the constant function on the integers $f(n) = 1$, is not a homomorphism with respect to addition, since it fails to map the identity element of the domain to the identity element of the codomain. But that text from Wikipedia appears to be using "codomain" to refer to what I'd call the "range" of a function, so a better example might be $g(n) = n+1$. Of course one can still "find" the structure of the domain in the structure of the codomain, even though it doesn't satisfy $g(n+m) = g(n) + g(m)$, but there are much "messier" permutations of $\mathbb{Z}$ if you prefer! $\endgroup$ – Steve Jessop Jul 10 '15 at 1:39
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Another term that is sometimes used, especially in the context of topological spaces and related objects, is "self-map".

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