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I was told that identity functions have the definition when followed.

If f ∘ e = f, e = idx. If e ∘ f = f, e = idy.

The textbook had a question below the definition asking, for any functions f and g, does either of g ∘ f = idx, f ∘ g = idy implies the other.

My answer was,

f ∘ g = idy

f ∘ g ∘ f = idy ∘ f = f

f ∘ (g ∘ f) = f

(g ∘ f) = idx (from the definition of f ∘ e = f, e = idx)

and similarly

g ∘ f = idx

f ∘ g ∘ f = f ∘ idx = f

(f ∘ g) ∘ f = f

(f ∘ g) = idy (from the definition of e ∘ f = f, e = idy)

But the answer was, no they do not imply the other and it didn't give me the proof, but showed a counter example.

counter example

Okay, I understand the counter example, but where in my proof has gone wrong? I didn't assume anything, just followed definition in the proof.

Edit: The text book says, (translated)

Let's figure out whether identity and inverse elements exist for composites. The identity element is a function e that satisfies, f ∘ e = e ∘ f = f for all f: X->Y. However, the function that satisfies f ∘ e = f is e: X->X, and the function that satisfies e ∘ f = f is e: Y->Y and hence they are different. Therefore, identity element of composites does not exist nor the inverse element. Despite this, we define the function e that satisfies f ∘ e = f as the identity function of X and write as id_X, and the function e that satisfies e ∘ f = f as the identity function of Y and write as id_Y. For each set, there exists one identity function.

Edit2: updated some lines for readability

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  • $\begingroup$ Who told you that definition? It's totally wrong... $\endgroup$ – Eric Wofsey Jul 14 '18 at 22:11
  • $\begingroup$ @EricWofsey in a textbook I am using. What is correct then? $\endgroup$ – Seung Jul 14 '18 at 22:23
  • $\begingroup$ If $f(x) = x^2$ and if $e(x) = -x$ then $f\circ e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $g\circ e(x) = e^{-x} \ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence. $\endgroup$ – fleablood Jul 14 '18 at 22:24
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    $\begingroup$ Usually the identy function is simply defined as $e: x\mapsto x$ for all $x$ in the domain. $\endgroup$ – fleablood Jul 14 '18 at 22:25
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    $\begingroup$ Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $f\circ e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context. $\endgroup$ – fleablood Jul 14 '18 at 22:36
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The problem is that you have no justification for jumping from$$f\circ(g\circ f)=f\implies g\circ f=\operatorname{id}_X,\tag1$$to $g\circ f=\operatorname{id}_X$. In fact, if $f$ was injective then, yes, you could deduce that $(1)$ hods, but all that you can deduce from $f\circ g=\operatorname{id}_Y$ is that $f$ is surjective, not that it is injective.

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  • $\begingroup$ f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx. $\endgroup$ – Seung Jul 14 '18 at 22:25
  • $\begingroup$ I have no idea about what you mean with “is in a form of”. Yes, $f\circ\operatorname{id}_X=f$, but what you asserted was that $f\circ(g\circ f)=f\implies g\circ f=\operatorname{id}_X$, without justification. $\endgroup$ – José Carlos Santos Jul 14 '18 at 22:30
  • $\begingroup$ Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here. $\endgroup$ – Seung Jul 14 '18 at 22:35
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    $\begingroup$ Yes, $\operatorname{id}_X$ is the only function from $X$ into itself such that for every $f\colon X\longrightarrow Y$, $f\circ\operatorname{id}_X=f$. But for a specific $f\colon X\longrightarrow Y$ there may be other functions from $X$ into itself (besides $\operatorname{id}_X$) with the same property. $\endgroup$ – José Carlos Santos Jul 14 '18 at 22:41
  • $\begingroup$ For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function? $\endgroup$ – Seung Jul 14 '18 at 23:04
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Okay.

You have two sets, $X$ and $Y$ and a class of functions that map $X \to Y$. They want to know if there is a function $id$ so that $id\circ f = f\circ id = f$ for any $f: X\to Y$.

I'm actually not sure why they even bothered to ask as the $id\circ f = f$ would require that $id: Y\to Y$ and $f\circ id = f$ would require $id:X\to X$ which is impossible as $X$ and $Y$ are different sets.

However then it asks if there are $id_X: X\to X$ that can act as a right identity with $f\circ id_X = f$ for all $f:X\to Y$. The answer is, yes, the identity element $i:X\to X$ where $i(x) = x$ for all $x \in X$ is such an element and with a little bit of playing it's clear that that is the is the only such function from $X \to X$.

(If $h(x) = w \ne x$ then there is some function $f$ where $f(w) \ne f(x)$ and so $f(h(x)) = f(w) \ne f(x) $. So the only function that works is $i(x) = x$.)

Likewise $i:Y\to Y$ $i(y) = y$ can act as $id_Y: Y\to Y$ that can act as a left identity with $id_Y\circ f = f$ for all $f: X\to Y$.

So they are asking if you have a $g:Y\to X$ so that $g\circ f: X\to X = id_X$ or $f\circ g: Y\to Y= id_Y$ does one imply the other.

And the answer is no.

The thing is if $|X| > |Y|$ then $f: X \to Y$ can not be injective so there must be $w\in X,x \in X$ and $w\ne x$ so that $f(x)=f(w)$. $g\circ f:X\to X$ can't be $id_x$. $id_X(w) = w$ and $id_X(x) = x$ but $g\circ f(w) = g(f(w)) = g(f(x)) = g\circ f(x)$.

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  • $\begingroup$ The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand. $\endgroup$ – Seung Jul 14 '18 at 23:37
  • $\begingroup$ I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property. $\endgroup$ – Seung Jul 14 '18 at 23:41

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