Is a continuous function with this property injective?

Question

Suppose that I have a continuous $f$ in a topological space with the following property:

for every continuous $g$ and $h$ such that $f \circ g = f \circ h$ $\Rightarrow$ $g=h$

It is true that a continuous $f$ with the property above must be necessarily injective?

I think that the easier approach is that one: One must show that if $f$ is not injective than it's possible to find continuous $g$ and $h$ such that $f \circ g = f \circ h$ but $g \neq h$

But I have no idea how to construct such $g$ and $h$. Any help would be appreciated

Answer 1

Let $X$ be the space, and suppose that $f(x_0)=f(x_1)$, where $x_0\ne x_1$. Let $$g:X\to X:x\mapsto x_0$$ and $$h:X\to X:x\mapsto x_1\;.$$