# Fundamental groups (isomorphism)

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.

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.