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Looking for input on, or alternative approaches to, my proof of the following statement:

Suppose $f$ is continuous on $(a,c)$. Further, for some $b$ satisfying $a \lt b\lt c$, let $f$ be strictly increasing on $(a,b)$ and strictly decreasing on $(b,c)$. Show that $f$ is not injective on $(a,c)$.

Firstly, by assumption, we must have that for any $x \in (a,b): f(b) \gt f(x)$. Similarly, for any $y \in (b,c): f(b) \gt f(y) \quad (*)$. To see this, consider if $f(b) \leq f(x)$ for some $x \in (a,b)$. This will lead to the contradiction of $f$ not being a strictly increasing function on $(a,b)$ (you can see how this contradiction will similarly arise for $f$ on the interval $(b,c)$ if we assumed $f(b) \leq f(y)$ for some $y \in (b,c)$).

Next, choose an $x \in (a,b)$ and choose a $y \in (b,c)$. Consider the intervals $[x,b]$ and $[b,y]$. $f$ is continous on each of these intervals, which means that there must be a minimum on each of these intervals. By $(*)$, we know that $f$ is strictly increasing on $[x,b]$ and strictly decreasing on $[b,y]$. Therefore, the minimum point in $[x,b]$ must be $x$, and the minimum point in $[b,y]$ must be $y$. If $f(x)=f(y)$, we are done. So suppose $f(x) \neq f(y)$. Then either $f(x) \lt f(y)$ or $f(y) \lt f(x)$. WLOG (without loss of generality), suppose the latter.

By $(*)$, we must have $f(b) \gt f(y) \gt f(x)$. Then, by applying the Intermediate Value Theorem to $[x,b]$, we know that there must exist a $z \in (x,b)$ such that $f(z)=f(y)$. Clearly, $z \neq y$ but $f(z)=f(y)$. Therefore, $f$ is not injective on $(a,c)$.

S.C.
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    Seems completely reasonable to me - mind if I ask exactly what WLG stands for? I assume it indicates that similar arguments work for both choices – Stephen Donovan Apr 16 '22 at 20:10
  • WLG = Without Loss of Generality. – Peluso Apr 16 '22 at 20:10
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    I usually see that written WLOG. – copper.hat Apr 16 '22 at 20:10
  • I agree, WLOG is a more common abbreviation, but I guess one understands from the context what he meant. – Peluso Apr 16 '22 at 20:12
  • @copper.hat roger that. I'll adopt that convention. Thank you – S.C. Apr 16 '22 at 20:13
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    I would note that $f(b)$ is the $\max$ on $(a,c)$ and pick some $y \in (\max(f(a),f(c)), f(b)$ and note that there is some $x_1 \in (a,b), x_2 \in (b,c)$ such that $f(x_1) = f(x_2) = y$. – copper.hat Apr 16 '22 at 20:13
  • @copper.hat: $f$ is only defined on the open interval $(a, c)$. I think that is why the author introduced $x$ and $y$. – Martin R Apr 16 '22 at 20:15
  • Proof seems fine. Could probably make it shorter, but depends on how much detail you are expected to give in the proof. For me, the symbol "∧" convinces me enough to move onto the next question. – Adam Rubinson Apr 16 '22 at 20:22
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    Perhaps one should also justify why "strictly increasing on $[x,b)$" plus continuous implies "strictly increasing on $[x,b]$". I also wouldn't say that an interval is increasing—the function is increasing on that interval. Nevertheless a very good solution. – Greg Martin Apr 16 '22 at 20:25
  • @GregMartin thank you for the input. I think my "$(*)$" is the result of the of "If $f$ is strictly increasing on $(a,b)$ and $f$ is continuous at $b$, then $f$ is strictly increasing on $(a,b]$". A proof by contradiction would show that if $f(b) \leq f(x)$ for some $x \in (a,b)$, then $f$ is no longer a strictly increasing function on $(a,b)$. Would you agree? – S.C. Apr 16 '22 at 20:36
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    As a follow-up (always follow up whether your instructor requests it or not): Give an example of an injective function that satisfies these condititions without the assumption of continuity. – B. S. Thomson Apr 16 '22 at 21:06
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    @B.S.Thomson I think $$f(x)=\begin{cases}x &\text{ if } 0 \lt x \lt 1 \ -x+3 &\text{ if } 1 \leq x \lt 2\end{cases}$$ should work. – S.C. Apr 16 '22 at 21:14
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    @S.Cramer Perfect. One more step. Characterize! Theorem. A continuous function $f:[a,b]\to\mathbb R$ is injective if and only if $f$ is either ___(?)___ or is ___(?)___. Prove that and you have beaten this problem to death, as it deserves. – B. S. Thomson Apr 16 '22 at 22:34
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    Assume $f$ is injective and apply https://math.stackexchange.com/questions/752073/continuous-injective-map-is-strictly-monotonic/1435015#1435015 – Michael Hoppe Apr 17 '22 at 19:08

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