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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.


(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.


(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.

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    $\begingroup$ What is $f\cup g$? $\endgroup$
    – mfl
    Commented Sep 26, 2018 at 16:30
  • $\begingroup$ @mfl I have given link of the book's page having the given questions. It is the union of two functions $f,g$. $\endgroup$
    – jiten
    Commented Sep 26, 2018 at 16:32
  • $\begingroup$ I have seen the link. What is the union of two functions? $\endgroup$
    – mfl
    Commented Sep 26, 2018 at 16:35
  • $\begingroup$ @mfl union of the mappings given their respective domains & ranges. $\endgroup$
    – jiten
    Commented Sep 26, 2018 at 16:36
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    $\begingroup$ We have $f\cup g:A\cup B\to C\cup D.$ How is it defined $(f\cup g)(a)$ for any $a\in A?$ $\endgroup$
    – mfl
    Commented Sep 26, 2018 at 16:39

1 Answer 1

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

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  • $\begingroup$ I request the calculus based approach for the part e, as shown in (ii) section under that. I am unable to continue it. $\endgroup$
    – jiten
    Commented Sep 26, 2018 at 16:50
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    $\begingroup$ In that case, you are no longer answering the original question. what are your domains of $f$ and $g$? $\endgroup$ Commented Sep 26, 2018 at 16:54
  • $\begingroup$ It is the default one, reals. Same for codomains. $\endgroup$
    – jiten
    Commented Sep 26, 2018 at 16:55
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    $\begingroup$ we are having the assumption that $f \cup g$ is a function. Since $A=C$ and $B=D$, we have $f=g$. $f \cup g = f$, so injectivity follows from injectivity of $f$. $\endgroup$ Commented Sep 26, 2018 at 17:00
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    $\begingroup$ To have $f \cup g$ remains a function, we can't afford to have $A=C$, $B=D$ and yet $f \ne g$. For the example that you claim $(0,0) \in f$, $(0,1) \in g$, $f \cup g$ is not a function. If you further impose the $f$ is continuous and injective, then we have $f$ is monotonic, hence $f=f \cup g$ is monotonic, hence it is monotonic and injective.... (monotonic is so restrictive, why are we even going there is puzzling to me) $\endgroup$ Commented Sep 26, 2018 at 17:20

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