If $f:R→R, g:R→R$ are continuous bijective, then prove/disprove $h(x)=f(x)−g(x)$ is a bijection function.

Question

*The full question is: *
If $f:R→R$ is a continuous increasing bijection function and $g:R→R$ is a continuous decreasing bijection function, then prove or disprove that $h(x)=f(x)−g(x)$ is a bijection function.

Am not good at analysis, know of the concept of continuity (from left & right, based on limits) only. This concept helps in deriving properties needed for my basic knowledge.
If two functions are monotonic (added to emphasize further the increasing/decreasing property), bijective (implies to me- that range exists for entire domain, but significance is not intuitive in this context) continuous functions, then even if the domain (& also range) of one is a subset (need not be proper), then still $h(x)$ 'will' not be bijective, as shown below:

If $f$ is suppose increasing, and $g$ decreasing over the same interval (both are having same domain and ranges, with none subset of the other, and are equal); then can imagine (with no value assigned to domain, function, hence to range to substantiate) that one function ($f$) is increasing with same domain and range, while another ($g$) is decreasing over the same domain (with the same range). Hence, the difference is zero, which forms the range of $h(x)$ and is a constant value, and hence is not a bijection function even, as $h(x)$ is having the same value for all elements. In case, the range elements have a constant difference for $f, g$, then still the same situation remains.

Answer 1

Guide:

• To prove that a fucntion is a bijection function, we have to check that they are both injection and surjection.

• To prove that it is an injection, suppose that $x_1 \ne x_2, h(x_1)=h(x_2)$, then we obtained $$f(x_1)-g(x_1)=f(x_2)-g(x_2)$$

$$f(x_1)-f(x_2)=g(x_1)-g(x_2)$$

Try to see a contradiction.

• To show surjection, note that $\lim_{x \to -\infty} f(x)-g(x)=- \infty$ and $\lim_{x \to \infty} f(x)-g(x)=\infty$, note that $f-g$ is also a continuous function and we can use intermediate value theorem on $h$ to show surjection.

Answer 2

The functions $f$ and $-g$ are both increasing and bijective. Therefore;

• $\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}-g(x)=+\infty$;
• $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}-g(x)=-\infty$.

So, $f-g$ is injective (since it is strictly increasing) and surjective ($\lim_{x\to\pm\infty}f(x)-g(x)=\pm\infty$ and surjectivity follws then from the intermediate value theorem).