Here is an assignment my professor handed out:
Let $X$ be some object crushed from a closed ball $B^2$.
Let $f:X\rightarrow X$ be a continuous function.
Then, there exists $x$ such that $f(x)=x$.
As you can see, the statement is really naive and informal.
I have no idea how to tackle this problem.. Please help
EDIT:
Actually, he simply drew this and asked us to prove $f$ has a fixed point
Let's suppose that by crushed, you're professor just means homeomorphic.
We suppose $X$ is homeomorphic to $B^2$ so there exists $j : B^2 \to X$ a homeomorphism, and $j^{-1} : X \to B^2$ is also a homeomorphism. We're given a continuous map $f : X \to X$, so let's consider the following chain of maps: $$B^2 \xrightarrow j X \xrightarrow f X \xrightarrow{j^{-1}} B^2$$
Define $h : B^2 \to B^2$ by $h = j^{-1} \circ f \circ j$ which is continuous since the composition of continuous maps is continuous.
By the Brouwer Fixed-Point Theorem, there exists $x \in B^2$ such that $h(x) = x$. In particular, $(j^{-1} \circ f \circ j)(x) = x$. But this means $f \big( j(x) \big) = j(x)$.
Hence $j(x)$ is the point in $X$ that satisfied the requirement you wanted.