MHB Solve Set Theory Question: Prove Iy o f = f

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The discussion revolves around proving the equality of the composition of functions, specifically showing that Iy o f = f, where f is a function from set X to set Y. The initial confusion arises from a potential misinterpretation of the function's domain and codomain, which is clarified by confirming that f should indeed be from X to Y. A straightforward proof is provided, demonstrating that for all x in X, the composition yields the original function f. The conversation also touches on the broader implications of function equality, emphasizing that even a single point of difference can invalidate equality, while also acknowledging contexts where functions can be considered equivalent under certain conditions. The discussion concludes with a note on the practical application of these concepts in calculus and measure theory.
Fernando Revilla
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I quote a question from Yahoo! Answers

Let f: A --> B. Prove that Iy o f = f
Here what I've got. Let, x is in X. Then there is a y in Y such that f(x) = y
=> Iy o f = Iy o f(x) = Iy(f(x)) = Iy(y). Please tell me what am I doing wrong in this question and how would you solve this? Thanks.

I have given a link to the topic there so the OP can see my response.
 
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I suppose you meant $f:X\to Y$ instead of $f:A\to B$. Then, simply:
$$\forall x\in X:\quad\left(I_Y\circ f\right)(x)=I_Y\left[f(x)\right]=f(x)$$
which implies $I_Y\circ f=f.$
 
Fernando Revilla said:
I suppose you meant $f:X\to Y$ instead of $f:A\to B$. Then, simply:
$$\forall x\in X:\quad\left(I_Y\circ f\right)(x)=I_Y\left[f(x)\right]=f(x)$$
which implies $I_Y\circ f=f.$
A nice simple little proof. Thank you!

-Dan
 
In general, when presented with two functions:

[math]f:A \to B[/math]
[math]g:A \to B[/math]

to decide whether or not the two functions are equal, we compare their values at every element [math]a \in A[/math]. In other words, we check if, for all such [math]a[/math]:

[math]f(a) = g(a)[/math] in [math]B[/math].

Even a single point of difference is enough to destroy the equality, as in, for example:

[math]f(x) = \frac{x}{x}; x \neq 0, f(0) = 0[/math]
[math]g(x) = 1[/math]

where the domain and co-domain of both functions are the real numbers.
 
Deveno said:
Even a single point of difference is enough to destroy the equality, as in, for example:

Right. However, I'd like to comment that sometimes we generalize the concept of function in some contexts. For example, for $(X,\mathcal{M},\mu)$ measure space and $f,g\in L^p(\mu)$ we need the equivalence relation $f\sim g$ iff $f=g$ almost at every point. So, we can define on the vector space $L^p(\mu)/\sim$ the norm $||f||_p=\left(\int_X|f|^p\right)^{1/p}$. We say that $f=g$ (in this context).
 
Well, sure. If two things aren't "quite equal enough" it's often common practice to "mod out the difference" and use equality of the resulting equivalence classes. It's sort of the raison d'etre of the notion of equivalence: all the properties of equality without the niggling details. For example: 2+2 and 4 are certainly not the same algebraic expression, but they have equivalent evaluations, which we use as if they were the same thing (through a process known as "substitution", or more generally, "representation").

Calculus students do this all the time when they find "the integral" of a function.
 
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