Mapping composition

Question

I have difficulties with this task: Prove, that every mapping can be written as a composition of surjective and injective mapping. So for example: $f \circ g = h$; $f$ is surjective; $g$ is injective; $h$ represents every mapping.

Additional task is whether changing the order of mappings ($g \circ f$) still became a mapping.

The thing i can prove and might be connected with the toppic is, that if $f\circ g =\text{identity}$, then $f$ is injective and $g$ is surjective.

Thank you very much!

Answer 1

You can always write a map $h:X\to Y$ as the concatenation of

• the restriction $h|_{\mathrm{im}(h)}:X\to\mathrm{im}(h)$ (which is surjective), and
• the embedding $\iota:\mathrm{im}(h)\to Y$ (which is injective).

Here $\mathrm{im}(h)=\{h(x)\mid x\in X\}\subseteq Y$ denotes the image of $h$. It is not hard to see why $\iota\circ h|_{\mathrm{im}(h)}=h$ holds.

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The other way around is possible too. Let $h:X\to Y$ and define $M:=X\,\dot\cup\, Y$. Then define maps $g:X\to M$ and $f:M\to Y$ via

$$g(x)=x,\qquad f(x)=\begin{cases} h(x)&\text{for } x\in X\\ x&\text{for } x\in Y \end{cases}.$$

Then $h=f\circ g$ with injective $g$ and surjective $f$.

Answer 2

Let's say $f:A\to B$ and $g:B\to C$ are mappings. Then the product or composition is a mapping $gf: A\to C$ with $gf(a) = g(f(a))$. In general, you can't reverse the order, because $g$ operates from $B$ to $C$, but the domain of $f$ is $A$, so $fg$ isn't even defined.

If we have a mapping $f:A\to B$, we want to show that there exist injective function $h$ and surjective function $g$ such that either $f = hg$ or $f = gh$.

Denote the set $$f(A) = \{f(a) : a\in A\} $$

If we let $g :A\to f(A)\subseteq B$ be defined such that for every $a\in A$ $g(a) = f(a)$, then it is clear $g$ is surjective. Now we automatically obtain an injective function $h:f(A)\to B$ (how?)