Why does the identity map induce the identity homomorphism on π₁(X)?

[Mikaelochi]

TL;DR

Another problem from a topology course I took and never really got

So, this problem I sort of get conceptually but I don't know how I can possibly rewrite (idX)∗ : π1(X) → π1(X). Does this involve group theory? It's supposed to be simple but I honestly I don't see how. Again, any help is greatly appreciated. Thanks.

 

[pasmith]

If $f : X \to Y$ is continuous, then $f_{*} : \pi_1(X) \to \pi_1(Y)$ is defined by $$f_{*}(\gamma) : [0,1] \to Y : t \mapsto (f \circ \gamma)(t)$$ for each $\gamma \in \pi_1(X)$.

If $X = Y$ and $f$ is the identity on $X$, then $f_{*}(\gamma) = \gamma$ and $f_{*}$ is the identity on $\pi_1(X)$.

 

[soloenergy]

Hi there! I can understand your confusion with rewriting (idX)* : π1(X) → π1(X). It might help to think of it in terms of group theory. Remember that π1(X) is the fundamental group of X, which is a group of loops in X that can be composed and inverted. So (idX)* is simply the identity map on the fundamental group, which just takes each loop to itself. This may seem trivial, but it becomes important when studying homotopy and homotopy equivalence.

Now, to rewrite this, you can think of it as (idX)* : G → G, where G is the fundamental group of X. This is a homomorphism from G to itself, which means it preserves the group structure. In other words, it takes the composition and inversion of loops in G to the composition and inversion of loops in G. So, in a way, it is just a fancy way of saying that (idX)* is the identity map on the fundamental group.

I hope this helps! Let me know if you have any other questions.