Maps homotopic between convex subspace of $\mathbb{R}^n$ and topological space. Let $X \subseteq \mathbb{R}^n$ be a convex subspace and let $Y$ be a path connected topological space, i.e every two pair of points are in the same path component. Show that any two maps from $X$ to $Y$ must be homotopic.How should one proceed here?
 A: Can you show that any map must be homotopic to a particular constant map? If two maps, $f$ and $g$, are each homotopic to a constant map $h$ with image $y_0$, via $\phi_f$ and $\phi_g$, then they are homotopic to each other via $\phi_g^{-1}\circ\phi_f$.

Thinking about this more, I think it's slightly more complicated. First, fix $x_0\in X$. Since $X$ is convex, we can smoothly contract $f$ to a map whose image is the single point $f(x_0)$. We just use $\phi_f(x,t)=f((1-t)x+tx_0)$. Similarly, we can write down $\phi_g(x,t)=g(tx+(1-t)x_0)$, which is the same thing for $g$, but running the other way in time. Both of these maps exploit the convexity of $X$.
Finally, we use the path connectedness of $Y$ by joining $\phi_f$ and $\phi_g$ with the path between $f(x_0)$ and $g(x_0)$. I guess we can write that down by saying, let $\chi:I=[0,1]\to Y$ be a path with $\chi(0)=f(x_0)$ and $\chi(1)=g(x_0)$.
We can stick these all together as follows: 
$$\phi:X\times I\to Y:(x,t)\mapsto\cases{\phi_f(x,3t) & $0\le t\le \frac13$\\ \chi(3t-1) & $\frac13\le t\le \frac23$\\ \phi_g(x,3t-2) & $\frac23\le t\le 1$}$$
We verify that this is well-defined and continuous, because:


*

*$\phi(x,\frac13)=\phi_f(x,1)=f(x_0)=\chi(0)$

*$\phi(x,\frac23)=\chi(1)=g(x_0)=\phi_g(x,0)$.


We verify that it is the right homotopy by checking:


*

*$\phi(x,0)=\phi_f(x,0)=f(x)$

*$\phi(x,1)=\phi_g(x,1)=g(x)$.


The trick is that all of the contracting and such has to happen back in $X$, where we have structure that allows things like multiplication by $t$, because we don't know anything about $Y$ other than the presence of paths.
