Show that if $f$ and $g$ are surjective, then $f\cup g$ is surjective.

Question

In the text of 'Introduction to Mathematical structures and Proofs', by Larry Garstein; there are below sub-problems as part of Q. 14 in Ex.#3.3, as shown here. Have taken a few questions below:

(d) Show that if $f$ and $g$ are surjective, then $f\cup g$ is surjective.
(e) Show that if $f\cup g$ is injective, then $f$ and $g$ are injective.
(h) Give an example of functions where $f$ and $g$ are injective, but $f\cup g$ is not injective.

I present here my attempt using three approaches:
(i) logic based approach that uses set-representation for domain & range,
(ii) calculus based approach by taking example of functions representing $f,g, f\cup g$.

(d) Show that if $f$ and $g$ are surjective, then $f\cup g$ is surjective.

(i) Both $f,g$ have their co-domain = range; that means that if consider the union of set of members in range (whether finite, or infinite) will have at least one member of the combined domain (set formed by union of domain sets of both $f,g$) that is pre-image of that.
Also, being union of sets, the common elements of the two sets (combined set of domain, or combined set of range) are repeated only once.

(ii) Let there be a polynomial $p_f$ that represents function $f$, & similarly $p_g$. Each polynomial denotes a mapping. Being surjective for any polynomial function means that two criteria are satisfied, as below :
(a) The function is unbounded from above & below, i.e.
$$ \lim_{x\to +\infty} f(x) = +\infty, \text{ and } \lim_{x\to -\infty} f(x) = -\infty $$ (b) Then, it is an added requirement that the polynomial is an odd order one. For an even order polynomial, for any $y\in \mathbb{R}$, there is $M > 0$ such that $$ x > M \Rightarrow f(x) > y, \text{ and } x < -M \Rightarrow f(x) < y $$ Hence, by intermediate value theorem, there is $x_0 \in [-M,M]$ such that $f(x_0) = y$.

So, the union of two surjective functions too is.

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(e) Show that if $f\cup g$ is injective, then $f$ and $g$ are injective.

(i) If $f\cup g$ is injective, then there is an underlying assumption: having common elements eliminated from the set of domain, & range of union of the functions $f,g$. Also, each element in the domain of set $f\cup g$ maps to a unique element in the range.

On separating out the respective members of domains of $f,g$ with duplication of common elements, get the two functions still injective.

(ii) For being an injective function, the polynomial representing the function should be monotonic. This can be checked by verifying that the first derivative is not infinity anywhere.
Stuck on this part.

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(h) Give an example of functions where $f$ and $g$ are injective, but $f\cup g$ is not injective.

Need help as am unable to think such example. Particularly, am not clear how it is possible theoretically.

Answer 1

For part $d$ and $e$, the task is to prove a general statement and we shouldn't assume structures such as we can we are dealing with real numbers, continuous functions, or polynomials.

$f:A \to B$, $g: C \to D$.

From the assumption, $f \cup g: A \cup C \to B \cup D$.

• For part $d$:

If $f$ and $g$ are surjections. Take an element from $y \in B \cup D$. If $y \in B$, we can find a preimage in $A$, hence we can find a preimage in $A \cup C$. If $y \in D$, then we can find a preimage in $C$ and we can find a preimage in $A \cup C$.

Hence $f \cup g$ is a surjection.

• For part $e$:

If $f \cup g$ is an injection, let's verify that $f$ is an injection. Suppose it is not, then we can find $x,y \in A$, $x \ne y$ such that $f(x)=f(y)$. That is we have found $x,y \in A \cup C$, $x \ne y$ such that $f(x)=f(y)$, violating the injectivity of $f \cup g$.

For part $h$:

Just let $B=D=\{1\}$, $A=\{1\}$, $C=\{2\}$.