If $ f $ is injective and $ g $ is injective, then $ f \circ g $ is surjective.

Question

I can prove that if $ f $ and $ g $ are both injective, then $ f \circ g $ is injective, but I don’t know how to prove that $ f \circ g $ is surjective.

Answer 1

This is not true, let $f(n)=n+1$, $g(n)=n$ for $n\in\Bbb N$. (Where $f,g:\Bbb N\to\Bbb N$.)

Answer 2

Let $A = \{a\}$, let $B = \{b\}$, and let $C=\{c_1,c_2\}$.

Define $f:B\to C$ to be $f(b)=c_1$

Define $g:A\to B$ to be $f(a)=b$

Here we have $A\xrightarrow{g} B\xrightarrow{f} C$, $f$ and $g$ are both injective (why?)

But, there is no preimage to $c_2$ (neither in $f$ nor in $f\circ g$). What does that mean about surjectivity?

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Here is a diagram of what is going on.

Remember the definitions of injectivity and surjectivity:

A function, $f:X\to Y$ is injective (also called one-to-one) iff whenever $f(x_1)=f(x_2)$ it implies that $x_1=x_2$.

In terms of the picture, a function is injective whenever there is no element with two (or more) arrows pointing to it. You see above that each element has only one or no arrows pointing to it, so both functions are injective.

A function $f:X\to Y$ is surjective (also called onto) iff every $y\in Y$ has some $x$ such that $f(x)=y$. I.e. every element in the codomain has a preimage. I.e. the Range of $f$ is the same as the Codomain of $f$.

In terms of the picture, a function is surjective whenever all elements in the codomain have at least one arrow pointing to them. Here we see that $2$ does not have an arrow pointing to it, so it is not surjective.

Finish by noting what $f\circ g$ looks like and that $2$ still doesn't have a preimage.

Answer 3

To give a counterexample where neither of the injective functions is surjective:

Consider $f:\mathbb N\to\mathbb N, n\mapsto 2n$ and $g:\mathbb N\to\mathbb N, n\mapsto 3n$. Then you get $f\circ g:\mathbb N\to\mathbb N, n\mapsto 6n$ which clearly isn't surjective; for example, there's no $n\in\mathbb N$ such that $6n=1$.