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Set theory question

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $I_Y\circ f=f.$
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
In general, when presented with two functions:

\(\displaystyle f:A \to B\)
\(\displaystyle g:A \to B\)

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

\(\displaystyle f(a) = g(a)\) in \(\displaystyle B\).

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

\(\displaystyle f(x) = \frac{x}{x}; x \neq 0, f(0) = 0\)
\(\displaystyle g(x) = 1\)

where the domain and co-domain of both functions are the real numbers.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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).
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.