in Set Theory & Algebra
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Justify the statements.

1. if f and f o g are one to one,does it follows that g is one to one.

2 if f and f o g are onto,does it follow that g is onto
in Set Theory & Algebra

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Statement 1 is Correct but Statement 2 is Incorrect.

Statement 1 : If $f$ and $f _o g$ are one to one, then it follows that $g$ is one to one.

Moreover, We don't even need $f$ to be One One. i.e. We have the following result :

If $f_og$ is One One then $g$ is One One.

Proof : 

We can use proof by Contradiction here.

So, Given that $f$ and $f_og$ are One to One. Let's assume that $g$ is Not One to one.

Since $f_og$ is One One, so, $(f_og(a) = f_og(b)) \rightarrow (a= b)$

Since $g$ is Not one to one, so, $\exists x,y (x\neq y \wedge (g(x) = g(y)))$ 

Since $g(x) = g(y)$, so, $f(g(x)) = f(g(y))$ i.e. $f_og(x) = f_og(y)$ which should mean that $x = y$ (because $f_og$ is One One) But we know that $x \neq y.$ Which is a Contradiction.

So, If $f_og$ is One One then $g$ is One One.  

Statement 2 : if $f$ and $f _o g$ are onto,  it follows that $g$ is onto. 

This statement is Wrong. We can give a Counter example for this statement :

Clearly, $f$ and $f_og$ are Onto But $g$ is Not Onto.

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1 comment

Thanks a lot sir!

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