Lifting a map to the total space of a circle bundle Let $\pi:P \to M $ be a smooth circle bundle, so $S^1$ is the fibre, $f:N \to M$ a smooth map. I would like to know what are the necessary and sufficient conditions for $f$ to lift to a map $g:N \to P$ covering $f$.
There are related questions reading which I think the answer is that $f$ lifts if and only if $f^*:H^2(M,\mathbb{Z})\to H^2(N,\mathbb{Z})$ is the zero map. However my understanding of algebraic topology is very basic and I cannot quite follow the arguments presented, so I would like for someone to confirm or  correct what I wrote. Incidentally, advice for the most accessible introduction to obstruction theory you know of would be appreciated.
In fact I am interested in a related but slightly different question: with $\pi:P\to M$ as above when does a diffeomorphism $f:M\to M$ lifts to a map $h:P\to P$? If the above criterium is correct $f$ lifts if and only if $\pi^*: H^2(M) \to H^2(P)$ is the zero map. Can anything be said about when that happens given that $P$ is a circle bundle?
 A: There are a few issues I'd like to address.  First, is a distinction that must be made between a circle bundle and a principal circle bundle.  What you described is a circle bundle: a fiber bundle where the fiber happens to be diffeomorphic to $S^1$.  In contrast, a principal circle bundle is a circle bundle for which there is a consistent choice of free,transitive $S^1$ action itself.
There are circle bundles which are not principal; the simplest example is probably the Klein bottle as an $S^1$ bundle over $S^1$.
A principal bundle over a base $B$ gives rise to a distinguished element $e$ (called the Euler class) which lives in $H^2(B;\mathbb{Z})$.  The condition on lifting is not that $f^\ast$ be the zero-map, but rather, that $f^\ast(e) = 0$.  Tsemo gave a fine proof that this condition is sufficient; the necessity follows from the fact that the kernel of the map $H^2(B;\mathbb{Z})\rightarrow H^2(P;\mathbb{Z})$ is generated by $e$.
But I want to stress that the argument involving the Euler class does not work for non-principal bundles (because the don't have an Euler class).  And, indeed, the result is false in this case.
To see this, consider the unit tangent bundle of $\mathbb{R}P^2$, $T^1\mathbb{R}P^2$.  This is diffeomorphic to the lens space $L(4,1)$, as argued in Proposition 4.2 of this paper.
Now, consider the natural projection $\pi:S^2\rightarrow \mathbb{R}P^2$.  Because $H^2(\mathbb{R}P^2;\mathbb{Z})$ is torsion, while $H^2(S^2;\mathbb{Z})$ is not, the induced map on $H^2$ must be the trivial map.
On the other hand, $\pi$ induces an isomorphism $\pi_2(S^2)\rightarrow \pi_2(\mathbb{R}P^2)$.  Because lens spaces are covered by $S^3$, they have trivial $\pi_2$, so the map $\pi:S^2\rightarrow \mathbb{R}P^2$ cannot lift to a map into $L(4,1)$.
A: You do not need obstruction theory for that: $f$ lift to $g$ if and only if $[v]=f^*([u])=0$ where $u\in H^2(M,\mathbb{Z})$ is the Euler class of $P\rightarrow M$. To see that:
Consider the pullback $f':P'\rightarrow N$ of $P\rightarrow M$. It is classified by $[v]$.  $[v]=0$, if and only if $P'\rightarrow N$ is trivial, This is equivalent to $P'$ is isomorphic to $N\times S^1$ which induces a section $g':N\rightarrow P'$ that can be composed with $P'\rightarrow P$ (the projection of the pullback). On the other hand, if the map $g$ is a lift of $f$, $g$ defines a section of the pullback of $P\rightarrow M$ by $f$, which is thus trivial since it is a principal bundle.
