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26 votes
26 votes

Let $f \: \circ \: g$ denote function composition such that $(f \circ g)(x) = f(g(x))$. Let $f: A \rightarrow B$ such that for all $g \: : \: B \rightarrow A$ and $h \: : \: B \rightarrow A$ we have $f \: \circ \: g = f \: \circ \: h \: \Rightarrow g = h$. Which of the following must be true?

  1. $f$ is onto (surjective)
  2. $f$ is one-to-one (injective)
  3. $f$ is both one-to-one and onto (bijective)
  4. the range of $f$ is finite
  5. the domain of $f$ is finite

4 Answers

Best answer
12 votes
12 votes

$f:A\rightarrow B$ is injective if and only if, given any functions $g,h:B\rightarrow A$ whenever $f\circ g=f\circ h,$ $f\circ g=f\circ h,then\ g=h$.

Refer to properties of Injective functions: https://en.wikipedia.org/wiki/Injective_function

Let us prove $(\forall g,h:f(g(x))=f(h(x))\rightarrow g(x)=h(x))\rightarrow f\ is\ one-to-one$ is true.

This is equivalent to, $f\ is\ not\ one-to-one\rightarrow (\exists g,h:f(g(x))=f(h(x))\wedge g(x)\neq h(x))$

Let us assume LHS is true, i.e. $f\ is\ not\ one-to-one$.

Then there exists some $c,d\in A$ such that,

                                     $f(c)=f(d)=a,$ where $a$ is an arbitrary element which belongs to B

Let g and h be some functions out of all possible functions from B to A such that $g\neq h$,

                                     i.e. $g(x)=c\ and\ h(x)=d\ \exists c,d\in A\ and\ \exists x\in B$

$\therefore f(g(x))=f(c)=a\ and\ f(h(x))=f(d)=a$ and $g(x)\neq h(x)$, i.e. RHS is also true.

Thus, whenever $\forall g,h\ f\circ g=f\circ h\rightarrow g=h$ is true, $f \ is\ one-to-one$.


Domain of $f$ need not be finite. Let $f:A\rightarrow B$ be identity function and A and B be infinite sets. Assume that $f\circ g=f\circ h$ is true,

then $f(g(x))=f(h(x))\rightarrow g(x)=h(x))$ will be true $\forall g,h$ since $f$ is an identity function. So, even if domain of f is not finite, the condition holds true.

Correct Answer: $B$

edited by
6 votes
6 votes
i think domain of F must be finite.
morover i think domain of F i.e, A shud be singleton set.
bcoz it is given that for all g and h possible between B and A, f∘g=f∘h⇒g=h..
if domain of F contains more than 1 element in domain,  f∘g=f∘h is possible but, g will not be equal to h
example:
suppose
A,B are set of all integers
g(x)= {x| for all integers x}
h(x)= {1 | for any integer x}
f(x) = {2| for any integer x}
if we see here,
f∘g=f∘h
f∘g = {(x,2)| for all integers x belongs to B}
f∘h = {(x,2) | for all integers x belongs to B}
but actually g and h are not equal.
so domain of f cant contain more than 1 element to ensure "if  f∘g=f∘h then g=h."
if A contains only one element, there is no other way for g and h except to map to that one element.
i mean number of functions possible from B-->A will be only one function if A is singleton set so, g and h will be equal obviously if f∘g=f∘h.
so i think E is the answer
5 votes
5 votes

We can tackle this question by elimination. 

| Domain | >= | Range |

So, if domain is finite, then range must be finite. both d and e must be true. So, domain is not finite since the question has a single option correct. 

Moreover, we can take an example to counter the statement that domain or range is finite. Example is -> 

Example 1-

if A = B = Set of all real no.s, and if g(x) = x-1, h(x) = x+1, f(x) = x (domain, range of all three functions not finite)

then the condition that f o g = f o h => g=h will also be satisfied. How?

The condition can be written as if g not equal h, then f o g not eq f o h.

In this case, let's assume f o g = f o h are equal even if g is not equal to h.

=>   x + 1 = x -1

=> 1 = -1 ( which is false always)

Hence, our assumption was wrong. Therefore, in this example the condition f o g = f o h => g=h is satisfied.

The options that remain are a,b,c.

Example 2-  

if A = Singleton Set, B = any set may or may not be singleton. From B -> A, only one function is possible because all the elements of the domain (B ) must be mapped to that single element in A only. Then, g = h is always true in this case. So, f o g = f o h => g=h will always be true. 

But, f from A -> B has a single element in the domain, so will be mapped to single element in B. Hence, for f is not onto. So, f is not a bijection also. Hence, option a and c are wrong. d and e were prooved wrong by the previous example.

So, the correct option is b. f must be one-one.

edited by
2 votes
2 votes

definition of one to one function --->

f(x1)=f(x2)===>x1=x2

fog(x)=f(g(x))

foh(x)=f(h(x))

now it is given that fog=foh==>g(x)=h(x)

then it prove from above definition is that it is a one -to-one functionhttps://en.wikipedia.org/wiki/Injective_function

check in definition part

Answer:

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