I work over an algebraically closed field of characteristic zero. Let $G$ be an algebraic group, $X,Y$ varieties with $G$-actions, and $\phi:X\to Y$ a $G$-equivariant morphism. Let $\mathcal{F}$ be a quasi-coherent $G$-equivariant sheaf of $X$. I want to show that $\phi_*\mathcal{F}$ has the natural structure of a $G$-equivariant sheaf.
Write $a_X:G\times X\to X$ and $a_Y:G\times Y\to Y$ for the action morphisms and $p_X:G\times X\to X$, $p_Y:G\times Y\to Y$ for the projections. Then since $\mathcal{F}$ is equivariant it is equipped with an isomorphism of sheaves $\varphi:a_X^*\mathcal{F}\to p_X^*\mathcal{F}$. Pushing this isomorphism forward along $(\text{id}_G\times\phi)$, we obtain an isomorphism of sheaves $$ (\text{id}_G\times\phi)a_X^*\mathcal{F}\cong (\text{id}_G\times\phi)p_X^*\mathcal{F} $$ My idea for obtaining an equivariant structure is to use a base change theorem applied to the squares $\require{AMScd}$ \begin{CD} G\times X @>{\text{id}_G\times\phi}>> G\times Y\\ @VVV @VVV\\ X @>{\phi}>> Y \end{CD} where the downward arrows are either $a_X,a_Y$ or $p_X,p_Y$. I think flat base change should apply here if the action morphisms and projection morphisms are flat. Of course the projection morphisms are flat, although I don't know if the action morphisms are flat but it seems like they should be. Is this correct? Does anyone have a reference?
You mean to show that the canonical map $$ a_{Y}^{\ast}(\phi_{\ast}\mathcal{F}) \to (\mathrm{id}_{G} \times \phi)_{\ast}a_{X}^{\ast}\mathcal{F} $$ is an isomorphism right? I think there is a commutative diagram $\require{AMScd}$ \begin{CD} G\times X @>{\alpha}>> G\times X\\ @V{a_{X}}VV @VV{p_{X}}V\\ X @>>{\mathrm{id}_{X}}> X \end{CD} where $\alpha : G \times X \to G \times X$ sends $(g,x) \mapsto (g,gx)$. This map $\alpha$ is an isomorphism because the map $(g,x) \mapsto (g,g^{-1}x)$ is its inverse. Thus $a_{X}$ and $p_{X}$ have the same properties.