# Morphism and Composing morphisms of Varieties

Hi guys I am trying to convince myself that composition of morphisms is again a morphism. If $\phi: V \rightarrow W$ and $\psi : W \rightarrow Z$ are morphisms of varieties. Then $\psi \circ \phi : V \rightarrow Z$ is a morphism and $(\psi \circ \phi)^* = \phi^* \circ \psi ^*$

I think the first part is easy to show something is a morphism we just need to show the image of the variety is in the second variety. Thus let $a \in V$ by $\phi$ been morpism $\phi (a) \in W$ now by $\psi$ been a morphism $\psi (\phi(a)) \in Z$ this we have shown the first part. For the second part I was thinking. Say $V \subset K^n \rightarrow W \subset K^m \rightarrow Z \subset K^p$ where we define $\phi (x_1,..x_n)= (f_1,...f_m)$ and $f_i \in K[x_1,..,x_n]$ and $\psi (y_1,...,y_m)= (g_1,...,g_p)$ and $g_i \in K[y_1,..,y_m]$ so now consider: $(\psi \circ \phi)^*=h \circ (\psi \circ \phi)=h(\psi(f_1,..f_m)=h(g_1(f_1,..f_m),..g_p(f_1,..f_m))$ but the problem is when I start to expand $\phi ^* \circ \psi ^*=h_1(\phi) \circ h_2(\psi)=h_1(f_1,..f_m) \circ h_2(g_1,..,g_p)$ and this is where I am stuck inputs are appreciated. Also a side question I was reading in the book and it said it is "To check that it is morphism, we need to check that the pullback maps zero to zero" I am confused why is that enough

• What you've shown is that the composition is a morphism of sets. There's more to show that the composition is a morphism of varieties. There are several equivalent ways to define a morphism of varieties, so it would be good to know which definition your book is using. Apr 20, 2015 at 4:15
• Sure the book definition is:
– Kori
Apr 20, 2015 at 4:22
• Sure the book definition is: Let $V \subset K^n$ and $W \subset K^m$ be affine varieties. A morphism $\phi : V \rightarrow W$ is a morphism from V such that $\phi (V) \subset W$. i.e. ther are polynomials $f_1,..f_m \in K[x_1,..x_n]$ where $\phi (a_1,..a_n)=(f_1(a_1,..a_n),...f_m(a_1,..a_n)) \in W$ for all $(a_1,..a_n) \in V$
– Kori
Apr 20, 2015 at 4:28

For the pullback, your use of the definition of the pullback is not quite right. What you wrote doesn't actually make sense, in fact, which is basically what you discovered. Instead, you need to apply the pullback to a function, so $$\left(\phi^* \circ \psi^*\right)h = \phi^*\left(\psi^*h\right) =\phi^*\left(h\circ\psi\right) =h\circ\psi\circ\phi =h\circ\left(\psi\circ\phi\right) =\left(\psi\circ\phi\right)^*h$$
Since that's true for arbitrary $h$, it shows that $\phi^*\circ\psi^*=\left(\psi\circ\phi\right)^*$
I'm not sure about the comment from the book, but I have an idea. Consider the map from the line $X$ with coordinate $x$ to the plane $Y$ with coordinates $y,z$ that sends a point $a$ to $(a,1)$. Then the pullback of the function $z-1$ is the zero function, even though $z-1$ is not zero itself, so the pullback doesn't send only zero to zero.