General situation
I'm preparing a geometry exam, and a lot of exercises from past years' exams are of the form «given $f$ the map so-and-so, prove (or determine whether) it is a homotopy equivalence». Most of the time, I'm pretty at a loss.
One possible trick (correct?)
Either the arrival space is contractible, from which I conclude any map from it to itself is homotopic to every other one (is that correct?), so I just need a putative homotopic inverse which is usually not too hard to find, or I start trying to think without too much success.
Another tool to be proved valid
One thing I would like to prove is that if two maps are homotopic, their images must have the same number of connected components. Suppose $f,g:X\to X$ are continuous and homotopic, i.e. there is $H:X\times[0,1]\to X$ a continuous map such that $H(x,0)=f(x),H(x,1)=g(x)$ for all $x\in X$. Suppose $f(X)$ is connected and $g(X)$ isn't. If I prove this leads to a contradiction, I will have proved my claim, since if $f(X)$ has $n$ connected components and $g(X)$ has $m<n$, I can restrict myself to the preimage of one connected component of $g(X)$ such that the image of that under $f$ is not connected, which must exist, and reapply the reasoning to their restrictions, which must still be homotopic. And anyway for now I have only needed this less general version. But how do I prove it? Let $x,y\in X$. Suppose $f(x),f(y)$ are in different connected components, whereas $g(x),g(y)$ are in the same. In particular, in the same pathwise connected component. I therefore have a curve from $g(x)$ to $g(y)$, let it be $\gamma$. This, I fear, won't bring me anywhere, as I can't make $\gamma$ "talk" with $f$ in anyway. Maybe we can try to restrict $H$ to the curve $\gamma$, but we'd have to bring it back to $X$ through a probably nonexistent inverse of $g$. Or maybe we can consider $K:[0,1]^2\to X$ defined as $K(t,s)=H(\gamma(t),s)$. But I can't see any use of this. Is my claim true? How do I prove it?
General question
Is there any theorem that could be of help, i.e. any theorem saying «if so-and-so, then the map is a homotopy equivalence»?
Particular case
In particular, I'm dealing with $V=\mathbb{CP}^2\smallsetminus\{X=0\}$, the projective complex plane being parametrized by $[X:Y:Z]$, and $\pi:V\to\mathbb{CP}^1$ the projection on the last two components $\pi([X:Y:Z])=(Y:Z)$. I have found a possible homotopy inverse $g([Y:Z])=[1:Y:Z]$. This is indeed a homotopy equivalence, and a homeomorphism with inverse $g^{-1}([X:Y:Z])=[\frac{Y}{X}:\frac{Z}{X}]$. $f\circ g([Y:Z])$ is clearly the identity. The other composition is $[X:Y:Z]\mapsto[1:Y:Z]$. How do I prove this is homotopic to the identity of $\mathbb{CP}^2$?