# Problem with path connectedness in intersection number $\operatorname {mod} 2$

An exercise from Guillemin-Pollack:

If $f\colon X\to Y$ is homotopic to a constant map, show that $I_2(f,Z)=0$ for all complementary dimensional closed $Z$ in $Y$, except parhaps if $\dim X=0$. [HINT: Show that if $\dim Z < \dim Y$, then $f$ is homotopic to a constant $X\rightarrow \{y\}$, where $y\notin Z$. If $X$ is one point, for which $Z$ will $I_2(f,Z)\ne 0$?]

Here $I_2(f,Z)$ is the intersection number $\operatorname {mod} 2$, i.e. the cardinality of $g^{-1}(Z)$ $\operatorname {mod} 2$ where $g$ is any map homotopic to $f$ and transversal to $Z$.

The only way of showing that $f$ is homotopic to a constant map $X\rightarrow \{y\}$, where $y\notin Z$ is by connecting a point in $Z$ with a point outside of $Z$ (i.e. in $Y\setminus Z)$ (which exists by dimension comparison) by a path. But nobody guaranteed that $Y$ is path connected. How to resolve this problem? Should I look for a different approach to this problem?

If $\dim Z<\dim Y$, then $Z$ not only is a proper subset of $Y$ but also cannot contain any component of $Y$. So $f$ is homotopic to some constant map with value $y$ (say), and there is a point $y'$ in the same component as $y$ which is not in $Z$ and then you can homotope $f$ along a path from $y$ to $y'$.
• How can I see that $Z$ cannot contain any component (I assume you mean connected component) of $Y$? Feb 27, 2018 at 0:36
• Every connected component of $Y$ is a submanifold of the same dimension as $Y$. Feb 27, 2018 at 0:37
• Now I'm confused by the fact that $I_2(f,Z)=0$ when $f$ is homotopic to the constant map at $y\notin Z$. Why is that so? I thought that since homotopic maps have the same $I_2$, the preimage of $Z$ under the above constant map is empty so $I_2=0$ by definition, but to apply the definition we also need that the above constant map is transversal to $Z$ (which isn't necessarily true I guess), so my argument is invalid? (Sorry, this actually is irrelevant to your answer, probably I need to ask it as a separate question.) Feb 27, 2018 at 19:24