In algebraic topology, for a function $f$ what does $f _\ast$ mean? In algebraic topology, for a function $f$ what does $f_{ \ast}$ mean? 
I'm solving some exercises and this is something that's appearing, often relating to homotopic functions, and I'm not sure what it means. Does anyone know?
 A: John has a good question. There is a tendency in algebraic topology to confuse a topological space and a topological space with a base point. Grothendieck wrote to me in 1983 in part: " both the choice of a base point, and the 0-connectedness assumption, however innocuous they may seem at first sight, seem to me of a very essential nature. To make an analogy, it would be just impossible to work at ease with algebraic varieties, say, if sticking from the outset (as had been customary for a long time) to varieties which are supposed to be connected. Fixing one point, in this respect (which wouldn't have occurred in the context of algebraic geometry) looks still worse, as far as limiting elbow-freedom goes!" But if $f:X \to Y$, then   $f_*$ could be a morphism $\pi_1(X,x) \to \pi_1(Y,f(x))$, or $H_n(X) \to H_n(Y)$.
A: If you have a continuous function between topological spaces (with base points)
$$
f: (X, x_0) \to (Y, y_0)
$$
then you have the induced function between the fundamental groups:
$$
f_* : \pi_1(X, x_0) \to \pi_1(Y, y_0).
$$
This function is defined by 
$$
f_*([\lambda]) = [f\circ \lambda].
$$
The $\circ$ is composition of functions.
Recall that the fundamental group is a set of (continuous) functions (loops) (the $\lambda's$) modulo homotopy. That is you have an equivalence relation on the set (of continuous functions)
$$
\{\lambda: [0,1] \to X: \lambda(0) = x_0 = \lambda(1)\}
$$
The equivalence relation is that $f\simeq g$ is there is a homotopy taking $f$ to $g$. When you mod out by this, you end up with a group (the fundamental group) where the elements are equivalence classes. Above, the induced function is defined on a representative, so you need to check that the definition is well-defined. That is, you need to consider what happens if you pick another representative.
That the above is well-defined comes from the fact that if $f\simeq f'$ and $g\simeq g'$ via homotopies, then $f\circ g \simeq f'\circ g'$ via the composition of the homotopies.
