# Reduced homology group of wedge sum

Let $X$ be Hausdorff space and let $x \in X$ such that there is a closed neighborhood $U$ such that $\{x\}$ is a strong deformation retract of $U$. If $Y$ is another Hausdorff space, I would like to show the inclusion maps induce isomorphisms

$$\tilde{H}_p(X) \oplus \tilde{H}_p(Y) \to \tilde{H}_p(X \vee Y)$$

where $X \vee Y = X \times \{y\} \cup \{x\} \times Y$ for some $y \in Y$. Since $H$ could be any homology theory, I need to derive this directly from the Eilenberg - Steenrod axioms. I probably need to use the additivity axiom which states that for a space $X = \coprod X_i$, the inclusion maps induce an isomorphism

$$\bigoplus H_n(X_i) \to H_n(X)$$

However, I'm not sure how to relate $X \coprod Y$ with $X \vee Y$. I would to solve this problem without using Mayer - Vietoris. Any help would be appreciated.

• Be careful, I think that -exactly- the point is not to use additivity. The additivity axiom (for a reduced theory) says exactly what you're trying to prove. The point here is that additivity for finite wedge is included in the Eilenberg-Steenrod axioms (not so for infinite wedges). – Bruno Stonek Jan 13 '16 at 10:01

Consider a pair $(X \lor Y, X)$. Let $Z = X - U$. By excision axiom, inclusion $(X \lor Y - Z, X - Z) \subset (X \lor Y, X)$ induces isomorphism in homology. As we have strong deformation retraction of $U$ on $x_0$, and since $(X \lor Y - Z, X - Z) = (U \lor Y, U)$, we get a deformation retraction of a pair $(U \lor Y, U) \to (Y, x_0)$, and the homotopy axiom says that this induces isomorphism in homology. The inverse is given by the map given by inclusion. Overall, we have that inclusion $(Y, x_0) \to (X \lor Y, X)$ induces an isomorphism. This map has an obvious left inverse $(X \lor Y, X) \to (Y, x_0)$ that contracts $X$ to a point. By easy categorical considerations, this map also must induce an isomorphism in homology.

Now consider the long exact sequence for pair $(X \lor Y, X)$:

$\cdots \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to \tilde{H}_{n-1}(X) \to \cdots$

The map $\tilde{H}_n(X) \to \tilde{H}_n(X \lor Y)$ is induced by inclusion $(X, x_0) \to (X \lor Y, x_0)$. There's an obvious left inverse to this inclusion, a map $(X \lor Y, x_0) \to (X, x_0)$ that contracts $Y$ in $X \lor Y$ to a point. As this map has a left inverse, the map induced in homology also has a left inverse, so it's a monomorphism, and we have an exact sequence $0 \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y)$

Similarly, consider $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$. As this map is also induced by inclusion $(X \lor Y, x_0) \to (X \lor Y, X)$, and from previous consideration we know that the contraction $(X \lor Y, X) \to (Y, x_0)$ induces an isomorphism, we consider a composition of these two. This map has a right inverse given by inclusion $(Y, x_0) \to (X \lor Y, x_0)$, which induces a left inverse for a map $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(Y)$, which necessarily is an epimorphism. Now compose it with the inverse isomorphism $\tilde{H}_n(Y) \to \tilde{H}_n(X \lor Y, X)$ to obtain that $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$ is also epimorphism, and thus we get an exact sequence $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to 0$. Remember that we composed it with the isomorphism inverse to the one $\tilde{H}_n(X \lor Y, X) \to \tilde{H}_n(Y)$ we used earlier, so here in this exact sequence, the map $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$ is the same as in long exact sequence.

Finally, we connect these two exact sequences to get a short exact sequence:

$0 \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to 0$

As the first map is induced by inclusion and it has a right inverse we mentioned earlier, we obtain that this s.e.s. split, so $\tilde{H}_n(X \lor Y) \simeq \tilde{H}_n(X) \oplus \tilde{H}_n(X \lor Y, X) \simeq \tilde{H}_n(X) \oplus \tilde{H}_n(Y)$.

Note that additivity axiom was not used.