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​Let $S$, $T$, $U$ be three (non-empty) sets.

Let $f:S\rightarrow T$ be a function with domain $S$ and codomain $T$, and let $g:T\rightarrow U$ a function with domain $T$ and codomain $U$.

The composition of $g$ with $f$, denoted $g\circ f$, is a third function $h=​g\circ f:S\rightarrow U$ with domain $S$ and codomain $U$, defined by $h(s)=(g\circ f)(s)=g(f(s))$, and also often called the composite function (when $f$ and $g$ are understood).

That is, we first apply $f$ to $s$ to yield $f(s)$, and then apply $g$ to $t=f(s)$ to get $g(t)=g(f(s))$.

Notice that it is not necessary for $f$ to have image all of $T$ for $g\circ f$ to be defined.

You may know the saying

Vedi Napoli e poi muori!,


See Naples and die!.

The intention is not that Naples is so bad it will kill you, but that it's so beautiful you would not want to be anywhere else for the rest of your life. Oh, and yes, this usually refers to Naples, Italy, not Naples, Florida (also a nice place, though). For people who love Venice, Italy, this saying is often recycled as See Venice and die!

Consider the set $S=\{$ sees NI, sees NF, sees VI, sees VF, sees VLOS, sees VLOU$\}$ where $NI$ is Naples, Italy; $NF$ is Naples, Florida; $VI$ is Venice, Italy; $VF$ is Venice, Florida; $VLOS$ is Venice, Los Angeles; and $VLOU$ is Venice, Louisiana.

Let $M=\{$ marries in NI, marries in VI$\}$, and

$W=\{$writes a diary, writes a book, writes a movie$\}$

$D=\{$ dies in NF, dies in VF, dies in VLOS, dies in VLOU $\}$.

For any two of the above sets, $A$ and $B$, and any function $f:A\rightarrow B$, let $f(a)=b$ mean in words $a$ then $b$, where then​ implies $a$ precedes $b$ in time.

For example, if $f:S\rightarrow T$ has:

$$f({\rm sees\,NF})={\rm marries \, in \,NI}$$

this means sees Naples, Florida, ​then marries in Naples, Italy.

Which of the following is TRUE​?


For $g:S\rightarrow S$, $h:S\rightarrow S$, we always have $h\circ g=g\circ h$.


For certain $f:S\rightarrow M$, $g:M\rightarrow W$, $h:W\rightarrow D$ we can have $(h\circ g)\circ f\not=h\circ(g\circ f)$.


There is an $f:S\rightarrow D$ such that there is no $g:D\rightarrow S$ with $(g\circ f)(s)=s$.


There are functions $f:S\rightarrow W$, $g:W\rightarrow S$ with $f\circ g=g\circ f$.

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