on the one hand it is clear, that for a degree $d$ map $f: \mathbb{P_\mathbb{C}}^1 \to \mathbb{P_\mathbb{C}}^1$ induces multiplication by $d$ on cohomology. On the other hand, if I use Poincare duality, I get a commutative diagram:
$\require{AMScd}$ \begin{CD} H^2(\mathbb{P_\mathbb{C}}^1 )\ @>{f^\ast}>> H^2(\mathbb{P_\mathbb{C}}^1 )\\ @VVV @VVV\\ H_0(\mathbb{P_\mathbb{C}}^1 ) @<{f_\ast}<< H_0(\mathbb{P_\mathbb{C}}^1 ) \end{CD}
From this would follow, that $f^\ast$ is the identity, as $f_\ast$ is. How can I solve this contradiction?
$\newcommand{\P}{\mathbb{P_C}^1}$ What makes you think this diagram commutes ?
Let's look at what it does: let $[\P]$ denote the characteristic class of $\P$. Then the vertical arrows are $x\mapsto [\P]\cap x$, so if you start in the top left hand corner, and follows right-down-left, you get $x\mapsto f_*([\P]\cap f^*x)$, whereas if you just go down, you get $x\mapsto [\P]\cap x$.
But now, naturality of the cap-product means $f_*([\P]\cap f^*x)= f_*[\P] \cap x$ (see e.g. here).
Now these two maps have no reason to agree, and in fact $f_*[\P]= d[\P]$, so they don't, unless $d =1$, which is what we would want