The definition of nullhomotopic is that: a map $f:X\to Y$ is nullhomotopic if $f$ is homotopic to a constant map i.e. there exists a homotopy $H(x,t):X\times I\to Y$ s.t. $H(x,0)=f(x),H(x,1)=x\mapsto y_0$. However, I have trouble understanding its intuition behind it. From here, it said that "nullhomotopy means we can homotope the image of $f$ to a point". To my understanding, that means we can deform $Y$ continuously to a point. Is it correct? Here are two examples I know the proof but try to figure out the intuition:
- Any continuous map $f:X\to Y$ where $Y$ is contractible is nullhomotopic. (Proved by construction)
- $f:S^n\to S^1$ for $n>1$ is nullhomotopic. (Proved using covering space)
- Any continuous map $f:\mathbb{R}P^2\to S^1$ is nullhomotopic.
Even though I can write down the proof exactly, I have trouble understanding the intuitions behind them. Would someone like to explain it to me?
No, that is wrong. First of all the quoted sentence talks about image, $Y$ is not necessarily the image (or equivalently: $f$ is not necessarily surjective). But more importantly, even when $f$ is surjective this is wrong. This intuition is dangerous, because there are nullhomotopic maps with non-contractible images. In fact you can find such map for any non-contractible Peano space (a space that is an image of $[0,1]$, or equivalently compact, connected, locally connected metric space) $X$, e.g. any sphere $S^n$. By taking any surjective curve $[0,1]\to X$ and concatenating it with the inverse of itself.
A better intuition, in my opinion, is to think about homotopy as a path inside the space of continuous functions $C(X,Y)$. This is even formally true when $X$ is locally compact Hausdorff. With this approach "$f$ is nullhomotopic" simply means that there is a path from $f$ to some constant map.
So for example your point 1 would mean: if $Y$ is contractible then $C(X,Y)$ is path connected. Point 2 means $C(S^n,S^1)$ is path connected for $n>1$. Point 3 means $C(\mathbb{R}P^2,S^1)$ is path connected.