Question

Let$\mathbf{f}\mathbf{:}\mathbf{B}\mathbf{\to }\mathbf{C}$ and$\mathbf{g}\mathbf{:}\mathbf{C}\mathbf{\to }\mathbf{D}$ be functions. Prove:

(a) If f and g are injective, then$\mathrm{g}\circ \mathrm{f}:\mathrm{B}\to \mathrm{D}$ is injective.

(b) If f and g are surjective, then$\mathrm{g}\circ \mathrm{f}$is surjective.

Short answer

1. It is proved that is$\mathrm{g}\circ \mathrm{f}:\mathrm{B}\to \mathrm{D}$ inject.
2. It is proved that$\mathrm{g}\circ \mathrm{f}$ is surjective.

Step-by-step solution

Write the given data from the question.

The functions are$\mathrm{f}:\mathrm{B}\to \mathrm{C}$ and $\mathrm{g}:\mathrm{C}\to \mathrm{D}$.

Prove that g∘f:B→D is injective.

(a)

The function f and g are injective.

To prove the given statement, it is needed to prove that$\mathrm{g}\circ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\circ \mathrm{f}\left(\mathrm{y}\right)$

Provided that$\mathrm{g}\circ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\circ \mathrm{f}\left(\mathrm{y}\right)$ thendata-custom-editor="chemistry" $\mathrm{x}=\mathrm{y}$

Therefore,

$$\begin{gathered} \mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{y}\right)\right) \\ \mathrm{f}\left(\mathrm{m}\right)=\mathrm{x} \\ \mathrm{f}\left(\mathrm{n}\right)=\mathrm{y} \end{gathered}$$

As function$\mathrm{g}:\mathrm{C}\to \mathrm{D}$ is injective therefore data-custom-editor="chemistry" $\mathrm{g}\left(\mathrm{m}\right)=\mathrm{g}\left(\mathrm{n}\right)$, then data-custom-editor="chemistry" $\mathrm{m}=\mathrm{n}$.

data-custom-editor="chemistry" $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)$

As function$\mathrm{f}:\mathrm{B}\to \mathrm{C}$ is injective therefore $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)$, then $\mathrm{x}=\mathrm{y}$.

It is shown that $\mathrm{g}\circ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\circ \mathrm{f}\left(\mathrm{y}\right)$.

Hence, it is proved that$\mathrm{g}\circ \mathrm{f}:\mathrm{B}\to \mathrm{D}$ is inject.

Prove that g∘f is surjective.

(b)

The function f and g are surjective.

To prove that$\mathrm{g}\circ \mathrm{f}$ is surjective, it is needed to prove that$\forall \mathrm{d}\in \mathrm{D},3\mathrm{b}\in \mathrm{B}$ Such that

$\left(\mathrm{g}\circ \mathrm{f}\right)\mathrm{b}=\mathrm{d}$

Let assume d is event of D.

Since$\mathrm{g}:\mathrm{C}\to \mathrm{D}$ is surjective then $3\mathrm{c}\in \mathrm{C}$,$\mathrm{g}\left(\mathrm{c}\right)=\mathrm{d}$

Since$\mathrm{f}:\mathrm{B}\to \mathrm{C}$ is surjective then $3\mathrm{b}\in \mathrm{B}$, $\mathrm{g}\left(\mathrm{b}\right)=\mathrm{c}$

Therefore,

$$\begin{gathered} \mathrm{g}\circ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{b}\right)\right) \\ \mathrm{g}\circ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{c}\right) \\ \mathrm{g}\circ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{d} \end{gathered}$$

Hence it is proved that$\mathrm{g}\circ \mathrm{f}$ is surjective.