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Consider the function

$f:X\rightarrow Y$ and $g:Y\rightarrow Z$

then which of the following is incorrect ?

(A) If $f$ and $g$ both are injective then $gof :X\rightarrow Z$ is injective .

(B) If $f$ and $g$ both are surjective then $gof :X\rightarrow Z$ is surjective.

(C) If $gof: X\rightarrow Z$ is bijective then f is injective and g is surjective .

(D) None

My Approach : For the (A) part since both f and g are one - one then I thought of some functions and hence came to the conclusion that $gof$ will be one - one . But I want to know some good and convincing approach for this question ...

Please help

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  • $\begingroup$ Use the definitions of injectivity and surjectivity. E.g., for (A), let $x,y\in X$ such that $g(f(x))=g(f(y))$. Then by injectivity of $g$, it must be that $f(x)=f(y)$, but then by injectivity of $f$ it must be that $x=y$. $\endgroup$ – smcc Jul 29 '16 at 16:39
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(A) $x\neq y$ implies $f(x)\neq f(y)$ implies $g(f(x)) \neq f(g(y))$

(B) For $z\in Z$ there is $y\in Y$ with $g(y)=z$ and then $x\in X$ with $f(x)=y$

(C) If $g\circ f$ is bijective and $V=f(X)$ (need not be all of $Y$) then $g:V\rightarrow Z$ is injective (but need not be injective on all of $Y$). But clearly $g$ must be surjective (or else you can't reach all of $Z$) and $f$ injective (or else some $x_1\neq x_2$ would map to the same point).

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  • $\begingroup$ H. H. Rugh I am sorry , I did not understood. $\endgroup$ – RAJESH SHARMA Jul 29 '16 at 16:43
  • $\begingroup$ Which part, if not all? $\endgroup$ – H. H. Rugh Jul 29 '16 at 16:47
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    $\begingroup$ Please explain the (A) part $\endgroup$ – RAJESH SHARMA Jul 29 '16 at 16:55
  • $\begingroup$ (A) Injective means that distinct points have distinct images. So we should show that $x\neq y$ implies $g(f(x))\neq g(f(y))$. First step: As $f$ is injective $x\neq y \Rightarrow f(x)\neq f(y)$. Second step: As $g$ is injective, $f(x)\neq f(y) \Rightarrow g(f(x)) \neq g(f(y))$ and we are done. $\endgroup$ – H. H. Rugh Jul 29 '16 at 17:49
  • $\begingroup$ @h.h.rugh how could you say that g:V→Z is injective? It need not be injective $\endgroup$ – user355797 Jul 31 '16 at 13:20
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THE ANSWER IS PART (C) .BECAUSE g$o$f is bijective does implies f is injective. But g must be bijective to satisfy the condition that g $o $f is bijective.if g is not injective then $x_1$ and $x_2$ can have same image in g .I.e Although $y_1=f(x_1)$ not equal to$ y_2=f(x_2)$,there may possibility that $g(y_1)=g(y_2)$ which disproves the statement that g $o$f is bijective.

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