How to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$

Question

Can somebody please explain how to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$

Answer 1

I'll give a couple hints, and then a solution below.

Hint 1: Notice that $S^1\vee S^1$ and $S^1 \times S^1$ are Eilenberg-Maclane spaces $K(\mathbb{Z}*\mathbb{Z},1)$ and $K(\mathbb{Z}^2, 1)$ respectively.

Hint 2: The inclusion $S^1\vee S^1 \to S^1 \times S^1$ induces the abelianization map on $\pi_1$.

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Solution

Let's say $F$ is the homotopy fibre of this map (using your favourite definition), so that $F\to S^1 \vee S^1 \to S^1 \times S^1$ is equivalent to a fibration sequence, hence we get a long exact sequence of homotopy groups. By Hints 1 and 2, the long exact sequence of homotopy groups tells use that $F\simeq K(C, 1)$ where $C$ is the commutator subgroup of $\mathbb{Z}*\mathbb{Z}$. This determines the (weak?) homotopy type of $F$ since the Eilenberg-Maclane property determines the homotopy type of a CW complex (and every space is weakly equivalent to a CW complex).

Since there are many different definitions of a homotopy fibre, usually we're only concerned with determining its homotopy type. Explicit descriptions are often opaque and hard to work with, so hopefully this is what the person posing the problem had in mind.

Answer 2

In general, the homotopy fiber of the inclusion $X \vee Y \hookrightarrow X \times Y$ is homotopy equivalent to the join $\Omega X * \Omega Y$. To see this, note that one definition of the homotopy fiber is the pullback of the maps $X \vee Y \to X \times Y$ and $P(X \times Y) \to X \times Y$, which is $(\Omega X \times PY) \cup_{\Omega X \times \Omega Y} (PX \times \Omega Y) \simeq \Omega X * \Omega Y$.

In the case $X = Y = S^1$, we have $\Omega S^1 \simeq \mathbb{Z}$, so the homotopy fiber is homotopy equivalent to $\mathbb{Z} * \mathbb{Z}$, which is some infinite bouquet of circles.