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I've been considering problems of the following type:

Given certain topological spaces $X$ and $Y$ and a continuous function $f:X\to Y$, prove that $f$ is null-homotopic.

The cases studied so far could all be solved in two steps:

  1. The spaces $X$ and $Y$ were such that no nontrivial morphism from $\pi_1(X)$ to $\pi_1(Y)$ could exist. So $f_* = 0$.
  2. The topological space $X$ was connected and locally arc-connected and $Y$ admitted a contractible covering space $E$. With the lifting lemma I could factor $f$ through $E$ because $f_*(\pi_1(X)) = 0 \subseteq p_*(\pi_1(E))$.

The conclusion followed using the contractibility of $E$ to prove that $f$ must be null-homotopic.Examples can be seen here and here.

My question is what can we say about $f$ if we don't have the technical hypotheses needed for the lifting lemma or if we don't know that $Y$ has a contractible covering space. In particular, I'm interested in the latter case. More precisely, the question is:

Let $X$ and $Y$ be arc-connected topological spaces such that no nontrivial morphism from $\pi_1(X)$ to $\pi_1(Y)$ can exist. Suppose that $X$ is locally arc-connected. If $f:X\to Y$ is a continuous function, what can we say about it? Is $f$ null-homotopic?

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Pipicito
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1 Answers1

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Typically, you can't say much. For instance, if $X=Y=S^2$, then the fundamental groups are trivial, but the identity map $f:X\to Y$ is not null-homotopic (you can prove this using homology groups, which unlike the fundamental group are nontrivial for $S^2$). In general, coming up with computable invariants that can detect when a map is not null-homotopic (or more generally, when two maps are homotopic) is very hard, and is more or less what the entire vast field of algebraic topology is about. The fundamental group is just one of the most basic such invariants, and in most cases it is far too weak to distinguish maps on its own.

Eric Wofsey
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