MadeEasy Test Series: Set Theory & Algebra - Functions

Question

Let f: A → B and g: B → C denote two functions. Consider the following two statements:
S1 : If both f and g are injections then the composition function gof : A → C is an injection.
S2 : If the function gof : A → C is surjection and g is an injection then the function f is a surjection.

Which of the above statements are valid?

a) S1 only

b) S2 only

c) S1 and S2

d) None of these

Comments

Both statements are valid.

---

i want to know how 2nd is valid

---

For function $f$ consider Set $A=\{a,b\},B=\{\alpha,\beta\}$

and mapping

$f(a)=\alpha,f(b)=\beta$

Consider function $g$ and set $C=\{1,2\}$ 

$g(\alpha)=1,g(\beta)=2$

So, till now both g and f are both injective and surjective.

So, my $g \circ f$ is surjective.

Now, let's try to make f not surjective, by adding an element $\gamma$ in set B for which there is no pre-image in set A,

so now my Set $B=\{\alpha,\beta,\gamma\}$

Now, since $g$ is injective and assuming it to be a total function, I have to add an extra element in set C so that

$g(\gamma)=3$. So ,now my set $C=\{1,2,3\}$

Now, our functions look like this

Now, if I have to make $g \circ f$ to be surjective, I have to provide a mapping from Set A, to the element $\gamma$ of set B, Otherwise, in the function represented by $g \circ f$ from $A \rightarrow C$, you would have no mapping for 3 and hence it would not be surjective.

So, yes I cannot make f to be not surjective in this case.

So I will add a new element to Set A which is $c$ such that $f(c)=\gamma$

So, your second statement is true.

Answer 1

BOTH STATEMENTS ARE TRUE BECAUSE S1 SAYS BOTH F AND G ARE ONE-ONE WHICH MEANS EACH ELEMENT IN A HAS AN IMAGE IN B,LIKEWISE B HAS IMAGE IN C ,THAT RESULTS ELEMENT IN A TO HAVE IMAGE IN C

FOR 2ND STATEMENT, WHEN G IS ONE ONE AND GoF is surjection,we can replace B with C and Get all preimages of C in A resulting F to be Surejection.