is my proof wrong?

Question

I have to find two functions such that one is surjective and the composition of them is not,so I chose $I:\Bbb N \mapsto \Bbb N$, $I(x)=x$ and $j:\Bbb R \mapsto \Bbb R$, $j(x)=x$

Both are surjective and the composition of them is $l(j(x))$ which means the domain and codomain become $l \circ j:\Bbb R\mapsto \Bbb R$; however, this isn't a function because it has elements in the domain that can't be mapped to a natural.

To show It's not surjective I used the number 5.7, which $j(5.7)=5.7 \implies l(5.7)$ is not defined because it's not in the codomain

The composition is not a function, because j(x) returns a real number, and I(x) is not defined for real numbers that are anot natural — Francisco José Letterio, Nov 29 '17 at 15:43

score 1 · Accepted Answer · answered Nov 29 '17 at 15:44

1

In order for the composition $f\circ g$ to make sense, the domain of $f$ has to equal the codomain of $g$. That is, if you have $g:A\to B$ and $f:B\to C$ then you can define $f\circ g:A \to C$. In your case, the domain of $\ell$ is $\mathbb N$, while the codomain of $j$ is $\mathbb R$, so it is nonsensical to form the composition $\ell\circ j$.

Here's a hint: if $f$ is not surjective, then $f\circ g$ is also not surjective (prove this).

answered Nov 29 '17 at 15:44

Mike Earnest

75,930

Yeh thanks just wanted to some clarification that it wasn't a function – john Nov 29 '17 at 15:52
if g is not surjective, then is f o g also not surjective? – john Nov 29 '17 at 20:33
@john not necessarily; it is still possible that f will be surjective when restricted to the range of g. For example, let f(x)=tan x (setting f(x)=0 when tan x is undefined) and g(x) = (pi/2) sin x. The range of g is only -pi/2 to pi/2, but f still attains every real number on this interval. – Mike Earnest Nov 29 '17 at 21:36

score 0 · Answer 2 · answered Nov 29 '17 at 15:41

Since the range of each function is not in the domain of the other function you cannot compose them as you've done here. Consider instead a function from any set non-empty to $\{0\}$ and some other function from $\{0\}$ to a set with more than one element.

is my proof wrong?

2 Answers2