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Homology classes under maps
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What happens to homology classes under continuous functions?
Let's consider a few examples with some typical maps.
Suppose $f \colon X \rightarrow Y$ is a map and $c$ is a homology class in $X$. Let's use the notation $$f_*(c)$$ for the corresponding homology class in $Y$.
If $f \colon X \rightarrow X$ is the identity map, each homology class is mapped to itself: $$Id_*=Id.$$
If $f \colon X \rightarrow Y$ is a constant map, all $k$-homology classes of $X$ with $k>0$ collapse or, algebraically speaking, are mapped to $0$ by $f_*$. Meanwhile, all $0$-homology classes of $X$ are mapped to the same $0$-homology class of $Y$. So it's the identity for a connected space. Note: If there is a non-trivial $k$-class in $Y$ with $k>0$ (like in the sphere ${\bf S}^k$), it has no preimages under $f_*$.
More specifically, suppose $X = {\bf S}^1$ and $Y = {\bf S}^1$, $b$ is a point in $Y$, $f \colon X \rightarrow X$ is a map. Let $P$ and $P'$ be the $0$-classes and $c$ and $c'$ the $1$-classes in $X$ and $Y$ respectively.
Now, if $f$ is the identity, we have $$\begin{array}{l} f_*(P) = P', \\ f_*(c) = c'. \end{array}$$
If $f$ is constant, given by $f(x) = b$ for all $x$, we have $$\begin{array}{l} f_*(P) = P', \\ f_*(c) = 0. \end{array}$$
To explain the last identity, observe: as the $1$-cycle that represents $c$ is mapped by $f$ into $Y$, the result is still a $1$-cycle, but since it is located entirely inside $b$, it's homologous to $0$.
Now, suppose $f$ is a flip of the circle. Then $$f_*(c) = -c'.$$
A map $X = {\bf S}^1 \rightarrow Y = {\bf S}^1$ can be thought of as a circular rope, $X$, being fitted into a circular groove, $Y$. You can carefully transport the rope to the groove without disturbing its shape to get the identity map. Or you can compress it into a tight knot to get the constant map.
You can also turn the rope before placing it into the groove. The resulting map is very similar to the identity regardless of the degree of the turn. Indeed, then $$f_*(c) = c'.$$
If you wind the rope twice before placing it into the groove, you get: $$f_*(c) = 2c'.$$
In the all examples above we have $$f_*(P) = P'.$$
Let's consider an example that illustrates what else can happen to $0$-classes. Suppose $f \colon X = {A, B} \rightarrow Y = [A,B]$ is given by $f(A) = A, f(B) = B$. Now, even though $A$ and $B$ aren't homologous in $X$, they are in $Y$. So, $f(A) \sim f(B)$. In other words, $$f_*([A]) = f_*([B]) = [A] = [B].$$
Algebraically, $[A] + [B]$ collapses to $0$.
A more advanced example is the collapse of the torus $f \colon X = {\bf T}^2 \rightarrow Y = {\bf S}^1$. We choose one longitude $L$ (in red) and then move every point in the torus to its nearest point on $L$. Then the homology class of the longitudes $L$ collapses while that of the latitudes $l$ is mapped to the $1$-class of the circle: $$\begin{array}{l} f_*(L) = 0, \\ f_*(l) = c. \end{array}$$
Next, consider the inclusion of the circle in the plane $f \colon X = {\bf S}^1 \rightarrow Y = {\bf R}^2$ (or the disk ${\bf D}^2$).
Here, the $1$-cycle in ${\bf S}^1$ is mapped to the identical cycle in ${\bf R}^2$, but this one is "filled": it's the boundary of the disk. Hence it's homologous to $0$, so $$f_*(c) = 0.$$
For more applications see Extensions. Another area where homology maps are necessary is persistent homology.
To see how to build $f_*$ continue to Chain map, Simplicial map, and Homology map.