So in my question I'm only asking for a morphism of Hodge structure. My point is that a map which respects the torus action seems to be stronger than a map which only respects the filtration.

Well, I never talked about mixed Hodge structure but only pure Hodge structure so there is no Galois action and therefore no weight filtration

Well I'll try to resolve my confusion, any way thanks Donu.

So going back to my example we find that $F^k H_{C}=C$ if $k\leq 0$ and $0$ if $k>0$. Similarly we find that $H_{\mathbf{C}}'=\mathbf{C}$ if $k\leq 1$ and $0$ if $k>1$. Therefore for all $k$ we have that $F^k H_{\mathbf{C}}\subseteq F^k H_{\mathbf{C}}'$.

So you see I'm not asking that my map respects the torus action but simply that it respects the Hodge filtration which is somehow weaker

Well may be I'm confused but take $H_{Q}=Q$ and $H_{Q}'=Q$ with the identity map $\iota: H_Q\rightarrow H_{Q'}$. If you place $H_Q$ in degree $(0,0)$ and $H_{Q}'$ in degree $(1,1)$ then this respects the Hodge filtration, isn't ?

My comment was meant to @Ulrich

Well I have a counter-example to what you said take $H_{\mathbf{C}}=\mathbf{C}$ place in degree $(0,0)$ and take $H_{\mathbf{C}}=\mathbf{C}$ place in degree $(1,1)$.

Well, if you use Hironaka's desingularization theorem then I guess that you can express (using the Leray spectral sequence) the cohomology of $Y$ in terms of the cohomology of a compactification of $Y$ and of the normal crossing divisors. Is it what you have in mind ?

Hi Pete, I guess that I had to reprove for myself that if $D$ (a divisor defined over $K$) has degree $1$ on a rational curve then necessarily it is equivalent to an effective divisor of degree $1$ defined over $K$ was not completely obvious to me, but I agree that it is fairly straightforward once you think about it.

In any case, thanks MP for your solution, I don't mind to use Riemann-Roch.

Ok so your argument is kind of non trivial since here you are using Riemann-Roch over the field $K$ isn't it ?