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Say I have two functions $f$ and $g$, which are continuous over the interval $[a,b]$ and also have the same domain (edit: I meant range not domain) $D_{fg}$ over that interval:

$\forall y \in D_{fg}. \exists x_1, x_2 \in [a,b]. f(x_1) = y \land g(x_2) = y$

If this is the case, then $f$ and $g$ must intersect at at least one point. (This becomes obvious with a piece of paper. If you draw the one function as surjective with respect to $D_{fg}$ and continuous on the interval it is not possible to also draw the other as continuous and surjective with respect to $D_{fg}$ on the interval without the lines crossing)

So formally then:

$\forall y \in D_{fg}. \exists x_1, x_2 \in [a,b]. f(x_1) = y \land g(x_2) = y \implies \exists y \in D_{fg}. \exists x. f(x) = g(x) = y$

How do I prove this implication however?

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    Try looking at the function $h = f - g$ and applying the intermediate value theorem in some way. – 2017-01-09
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    @dxiv But a parabola over a symmetric interval centered around its extremum is also surjective and continuous but not monotonic, or what am I missing here? – 2017-01-09
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    You're missing that dxiv deleted that comment. I expect that (like me at first) dxiv saw the word "injective" when it wasn't actually there. :) – 2017-01-09
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    @user3578468 What John Hughes said ;-) – 2017-01-09
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    Peculiar terminology. A more usual term is range. Is the range a subset of the reals? – 2017-01-09
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    Domain is confusing here, I would use domain for input values. Instead "image" is a better word, do you mean that if $f([a,b])=g([a,b])$ then $\exists x\in [a,b]\,|\,f(x)=g(x)$ ? – 2017-01-09
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    @zwim yea I meant image – 2017-01-09

2 Answers 2

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By the extreme value theorem, there exist an $x_f$ such that $f(x_f) = \max\{f(x)\mid x \in [a,b]\}$ and an $x_g$ such that $g(x_g) = \max\{g(x)\mid x \in [a,b]\}$. Moreover, $f(x_f) = g(x_g) = c$. If $g(x_f) = c$ or $f(x_g) = c$, we are done. Otherwise, $g(x_f) < c = f(x_f)$ and $f(x_g) < c = g(x_g)$. Now apply the intermediate value theorem to $f - g$ on the interval $[x_f,x_g]$ or $[x_g,x_f]$.

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Since the two functions have the same range (I assume that is what you meant; otherwise the question is not true) then there are values $a \le c, d \le b$ with $c \ne d$ such that $f(c) \le f(x) \le f(d)$ and $f(c) \le g(x) \le f(d)$ for $a \le x \le b$.

I will assume that $c < d$. If $c > d$, just switch them in what follows.

Consider $h(x) = f(x)-g(x)$ for $c \le x \le d$.

Then $h(c) = f(c)-g(c) \le 0$ (since $f(c) \le g(x)$ for $c \le x \le d$) and $h(d) = f(d)-g(d) \ge 0$ (since $f(d) \ge g(x)$ for $c \le x \le d$).

Therefore there is a point $v$ such that $c \le v \le d$ and $0 = h(v) =f(v)-g(v) $ or $f(v) = g(v)$.