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If functions $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{g}: \mathrm{B} \rightarrow$ A satisfy gof $=\mathrm{I}_{\mathrm{A}}$, then show that $\mathrm{f}$ is one-one and $g$ is onto.
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Verified Answer
Given that,
$\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{g}: \mathrm{B} \rightarrow$ A satisfy $g \circ f=\mathrm{I}_{\mathrm{A}}$ $\because$ gof $=\mathrm{I}_{\mathrm{A}}$
$$
\begin{aligned}
&\because \operatorname{gof}=\mathrm{I}_{\mathrm{A}} \\
&\Rightarrow \mathrm{gof}\left\{\mathrm{f}\left(\mathrm{x}_1\right)\right\}=\mathrm{gof}\left\{\mathrm{f}\left(\mathrm{x}_2\right)\right\} \\
&\Rightarrow \mathrm{g}\left(\mathrm{x}_1\right)=\mathrm{g}\left(\mathrm{x}_2\right) \quad \quad\left[\because \mathrm{gof}=\mathrm{I}_{\mathrm{A}}\right] \\
&\therefore \mathrm{x}_1=\mathrm{x}_2
\end{aligned}
$$
Hence, $f$ is one-one and $g$ is onto.
$\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{g}: \mathrm{B} \rightarrow$ A satisfy $g \circ f=\mathrm{I}_{\mathrm{A}}$ $\because$ gof $=\mathrm{I}_{\mathrm{A}}$
$$
\begin{aligned}
&\because \operatorname{gof}=\mathrm{I}_{\mathrm{A}} \\
&\Rightarrow \mathrm{gof}\left\{\mathrm{f}\left(\mathrm{x}_1\right)\right\}=\mathrm{gof}\left\{\mathrm{f}\left(\mathrm{x}_2\right)\right\} \\
&\Rightarrow \mathrm{g}\left(\mathrm{x}_1\right)=\mathrm{g}\left(\mathrm{x}_2\right) \quad \quad\left[\because \mathrm{gof}=\mathrm{I}_{\mathrm{A}}\right] \\
&\therefore \mathrm{x}_1=\mathrm{x}_2
\end{aligned}
$$
Hence, $f$ is one-one and $g$ is onto.
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