Counterexamples on homotopy equivalence and infinite product

Question

1. Let $(X,a),(Y,b)$ be pointed spaces and $f:(X,a)\rightarrow (Y,b)$ be a continuous function. If the natural homomorphism $f_*:\pi_1(X,a)\rightarrow \pi_1(Y,b)$ is a group isomorphism, is $f$ necessarily a homotopy equivalence?

2. Let $\{(X_i,x_i)\}$ be an infinite collection of pointed spaces. Then, is $\pi_1(\prod X_i, \prod x_i) \cong \prod \pi_1( X_i, x_i)$? What is a counterexample? Moreover, if this is not true, then is the fundamental group of infinite product group-isomorphic to direct sum of fundamental groups?

Answer 1

1. This is not true, as explained in the comments: the inclusion map $f : \{pt\} \to S^n$ of a point induces the zero morphism $f_*$ at the level of $\pi_1$. Yet, $f$ is not a homotopy equivalence as $S^n$ is not contractible (which can be proved by computing homology). The reason this fails is that $\pi_1$ only detects homotopy $1$-type. One needs to look at the higher homotopy groups $\pi_n(X, x_0)$ for $n > 0$, defined to be homotopy classes of based maps $(S^n, pt) \to (X, x_0)$, where an operation similar to composition makes it into a group. Then the following is true, and is a hard theorem of Whitehead: If $X, Y$ are connected CW-complexes, $f : (X, x_0) \to (Y, y_0)$ is a based map inducing isomorphism $\pi_n(f) : \pi_n(X, x_0) \to \pi_n(Y, y_0)$ for all $n \geq 1$, then $f$ is a homotopy equivalence.

2. Yes, this is true. $(\prod X_i, \{x_i\})$ is product of $(X_i, x_i)$'s in $\mathsf{Top}_*$, and $\pi_1$ preserves all based products. In a concrete way, $X = \prod X_i$ has the coarsest topology in which the projection maps $p_i : X \to X_i$ are continuous. Thus, a path $f : I \to X$ is the "same" as collection of paths $p_i \circ f : I \to X_i$, and a homotopy $\{f_t : I \to X\}$ of paths is the "same" as a collection of homotopies $\{p \circ f_i : I \to X_i\}$ for each $i$. Thus, the bijection $\pi_1(X) \to \prod \pi_1(X_i)$ given by sending a loop $\gamma$ to the tuple of loops $(p_i \circ \gamma)$ is well-defined, and is a group homomorphism as it is so componentwise. We have our desired isomorphism.