Show that two continuous functions that are surjective over the same interval intersect 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?
 A: 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 $[x_g,x_f]$.
A: 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
$f(v) = g(v)$.
