# Prove that any two maps $f, g:X \to Y$ are homotopic

Let $X$ be a topological space and $Y$ be the upper open hemisphere. Prove that any two maps $f, g:X \to Y$ are homotopic.

I have just started to learn algebraic topology and just learn the definition and few example of homotopy.
Here I need to find a function $F:X \times I \to Y$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$
But I could not find such a function. Can somebody help me please.
Are there any general way to find a homotopic function?

• Hint: Prove than any $f:X\to Y$ is homotopic to a particular constant map. Then show that if $f$ is homotopic to $h$ and $h$ homotopic to $g$ then $f$ is homotopic to $g$. – Thomas Andrews Aug 7 '13 at 17:37
• @Thomas Andrews, will you please explain me how can I able to show that any $f:X→Y$ is homotopic to a particular constant map. – habul Aug 7 '13 at 17:45
• You really only need one such homotopy, with $X=Y$ and $f=\mathrm{Id}_Y$. See my answer for some details. I don't tell you how to construct that particular example, only how once you have such an example, you have it for all $f$. – Thomas Andrews Aug 7 '13 at 17:48

Note the upper open hemisphere is homeomorphic to the plane. Then just take $F(x,t)=tf(x)+(1-t)g(x).$

Take \begin{align*} F(x,t)=\frac{tf(x)+(1-t)g(x)}{\|tf(x)+(1-t)g(x)\|}. \end{align*}

Because $f(x)$ and $g(x)$ lie in the upper open hemisphere, $tf(x)+(1-t)g(x) \neq 0$ for all $x$ and $t$ so $F$ is well-defined.

• @ frankieiknarf1618 in the above given answer can you please tell me why do we need that denominator factor when the numerator part is already lying in the hemisphere – junegirl Aug 12 '13 at 15:38

Pick $y\in Y$ to be the apex of the hemisphere. (It actually doesn't matter which point you choose.) Define a homotopy between $c_y,\mathrm{id}_Y:Y\to Y$ where $c_y$ is the constant function taking $y$ as the value, and $\mathrm{id}_Y$ is the identity function on $Y$.

Once you have this homotopy, show that there is a homotopy between $f$ and $c_y\circ f$. Then show that $c_y\circ f = c_y\circ g$, so there is a homotopy between $f$ and $g$.

This shows how this result comes from a single homotopy between two specific functions in $Y\to Y$. A $Y$ which has a homotopy between $\mathrm{Id}_Y$ to a constant function is called "retractable to a point."