Question

If $\mathit{f}\mathbf{}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{C}$is a function, then f can be considered as a map from B to since$\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{\in }\mathit{I}\mathit{m}\mathit{f}$ for every . Show that the map$\mathit{f}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{I}\mathit{m}\mathit{f}$ is surjective.

Short answer

Thus, the map$f:B\to lmf$ is surjective.

Step-by-step solution

Showing that f[A]= lm (f) = B

Consider f is surjective, meaning that for each$b\in B$ there exists an $a\in A$such that b=f (a).

Consider, first that f is surjective for each $b\in B$, $\exists a\in A$such that b = f (a) .

Hence, $f\left[A\right]=lm\left(f\right)=B$.

Showing that f (a) = b

Conversely consider $f\left[A\right]=lm\left(f\right)=B$, then for eachrole="math" localid="1659173826544" $b\in B$ there exists and$a\in A$ such that $b=f\left(a\right)$, which is the definition of surjectivity.

Hence, the map$f:B\to lmf$ is surjective.