Is $f_1: \pi_1(X \vee X) \rightarrow \pi_1(X)$ a surjection?

Question

Suppose we have a space $X$ Does the following always hold?

Does there exist a map $f: \pi_1(X \vee X) \rightarrow \pi_1(X)$ which is surjective?

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More generally does

Do there exist maps $f_1: \pi_1(X \vee Y) \rightarrow \pi_1(X)$ and $f_2:\pi_1(X \vee Y) \rightarrow \pi_1(Y)$ which are surjective?

this hold true? I think the answer is yes by the SVK but I am not sure about it.

Answer 1

$X \vee Y$ is the one-point-union of the pointed spaces $X,Y$. It can represented as the quotient space

$$X \vee Y = (X \times \{ 1\} \cup Y \times \{ 2\}) / (\{x_0\} \times \{ 1\} \cup \{y_0\} \times \{ 2\} ) .$$

The map $i_X : X \to X \vee Y, i_X(x) = [x,1]$, is an embedding, similarly $i_Y : Y \to X \vee Y, i_Y(y) = [y,2]$.

Define $R_X : X \times \{ 1\} \cup Y \times \{ 2\} \to X, R_X(x,1) =x, R_Y(y) = x_0$. This induces a map $r_X: X \vee Y \to X$ on the quotient having the property $r_X \circ i_X = id$. Thus $r_X$ is a retraction. Similarly we get a retraction $r_Y : X \vee Y\to Y$.

Now let $f_X = (r_X)_* : \pi_1(X \vee Y) \to \pi_1(X)$. Then $f_X \circ (i_X)_* = id$ which shows that $f_X$ is a surjection. Simlarly $f_Y = (r_Y)_* : \pi_1(X \vee Y) \to \pi_1(Y)$ is a surjection.

Answer 2

This is much simpler than applying the Seifert-Van Kampen theorem.

If you set things up correctly it's just an application of the functoriality properties of the fundamental group functor: By folding $X \vee X$ in half one gets a retraction from $X \vee X$ to the first $X$ factor of $X \vee X$, and now you can apply the fact that the induced $\pi_1$-homomorphism of a retraction is surjective.

In more detail, pick a base point $p \in X$ and represent $X \vee X$ as $$X \vee X = (X \times \{p\}) \cup (\{p\} \times X) \subset X \times X $$ with base point $P = (p,p) \in X \vee X$. Define an embedding $$i : (X,p) \to (X \vee X,P) \quad i(x)=(x,p) $$ and a projection $$q : (X \vee X,P) \to (X,p) \quad q(x,y)=x $$ Intuitively one is embedding $X$ as the "first" factor of $X \vee X$ and then one is retracting $X \vee X$ to that free factor, although formally it is easier to just define and use the embedding.

Since $q \circ i : (X,p) \to (X,p)$ is the identity, it follows that the composed induced group homomorphism is the identity: $$(q \circ i)_* : \pi_1(X,p) \to \pi_1(X,p) $$ Since $\pi_1$ is a functor, this homomorphism factors as $$(q \circ i)_* : \pi_1(X,p) \xrightarrow{i*} \pi_1(X \vee X,P) \xrightarrow{q_*} \pi_1(X,p) $$ Since the composition $(q \circ i)_* = q_* \circ i_*$ is surjective, it follows that the second factor $q_*$ is surjective.