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Regarding the proof of the theorem that says the fundamental group of a product space is isomorphic to the product the fundamental groups of the two spaces. A more detailed proof that I manage to find made use of projections of the loop. The homomorphism defined in the proof is as follows:

$\phi:\pi_1(X\times Y,(x_0,y_0))\rightarrow \pi_1(X,x_0)\times\pi_1(Y,y_0)$ where $\phi([f])=([proj_1\circ f],[proj_2\circ f])$

and also the inverse homomorphism

$([f],[h])\rightarrow [(f,h)(t)]$ where $(f,h)(t)=(f(t),h(t))$.

It was also mentioned that two facts are required:

  1. If $f\simeq g:[0,1]\rightarrow X\times Y$, then $proj_i\circ f\simeq proj_i \circ g$ for $i=1,2$
  2. If we take $f\ast g:[0,1]\rightarrow X \times Y$, then $proj_i\circ(f\ast g)=(proj_i\circ f)\ast(proj_i\circ g)$

Below are a few doubts that I have for this proof and would really appreciate if anyone can clarify. I am a beginner in learning algebraic topology.

Question 1: Why do we need fact (1) in the proof? Correct me if I am wrong, the symbol "$\simeq$" refers to "homotopic" between two loops in this context am I right? Is it because when we define the homomorphism $\phi$ for two the two groups the input for $\phi$ must be an equivalence class?

Question 2: Fact (2) is required as mentioned in the proof. Is it because a group homomorphism must preserve the operations? Because that is what I see from most of the definition of group homomorphism.

Question 3: The proof merely claimed the two facts but I am not sure if it is too trivial to prove them. If I would like to prove, is it very tedious? Any hints or some guide would definitely help.

Thanks.

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Basically, you're right.

Fact (1) is needed to have a well-defined map. You want $[f]$ to have the same image no matter what representative of its class you take, and in this case $[f]$ is an homotopy clas, so the relation satisfied is to be homotopic. Since the image is also a homotopy class, then the images of two representatives of the same class have to be homotopic.

Fact (2) is required because a group homomorphism must respect the group operation. In the case of the fundamental group, the operation is the concatenation of loops modulo homotopy, so it has to take concatenations into concatenations (at least modulo homotopy, though you have indeed equality).

For your las question the answer is no, it's not very hard to prove those properties, which are usually included in any basic course of homotopy theory. Here is a prove for the first one (actually a more general fact), and the second is not hard, just write down the equations.

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