Prove: for any maps $f,g : \Sigma X \rightarrow Y$ , then $(f + g)_* = f_* + g_* : H_*(\Sigma X) \rightarrow H_*(Y )$. And prove the same for cohomology
I know how to prove the composition as here Proving $(f \circ g)_{*}= f_{*} \circ g_{*}$ , but here this map is from the homology of the suspension of the space not the space itself. could anyone help me in proving this, please?
The meaning of addition of $f,g$ in this context is:
As for adding functions: for any space X and any space Y, $[\Sigma X, Y ]$ is a group (because the suspension is a co-$H$ space). Therefore, if $f, g: \Sigma X \to Y$, then there is also their sum $f + g: \Sigma X \to Y.$
Explicitly, you pinch the equator of the suspension to get a wedge of two copies of $\Sigma X$, and then apply $f$ to the top wedge summand and $g$ to the bottom wedge summand.
This answer is adapted from Lemma 4.60 of Hatcher. The original argument there relies on chasing elements through diagrams. Here I present a more categorical version of the proof.
Note that I have deviated a little bit from Hatcher's notation by using both $\times$ and $\oplus$ in places where he uses $\oplus$ only, even though the direct sum and direct product of two summands/factors are the same. This is to differentiate between maps into vs. out of the direct product/sum, so that it's clear which limit/colimit diagram we are using, and makes it obvious what changes when we dualize.
As clarified by the discussion in the comments as well as your edit, the map $f + g$ is the composition $\Sigma X \xrightarrow{c} \Sigma X \vee \Sigma X \xrightarrow{f\vee g} Y$, where $c$ is the "pinch" map that quotients out an equatorial copy of $X$ in the suspension. Applying the homology functor to this, we get the map $(f+g)_*$. We want to prove that this map is in fact equal to $f_* + g_*$.
The key idea is to think of $f_* + g_*$ as the composition of the diagonal homomorphism $\Delta: H_\bullet(\Sigma X) \to H_\bullet(\Sigma X) \times H_\bullet(\Sigma X)$ given by $x\mapsto (x, x)$ with the direct sum $f_*\oplus g_*$. Now recall that $H_i(\Sigma X\vee \Sigma X) \cong H_i(\Sigma X) \oplus H_i(\Sigma X)$. Thus if we can show that the following diagram commutes, and the two vertical arrows cancel (i.e. $j$ is the inverse of $(i_{1*}\oplus i_{2*})$), then we will have obtained the equality between the composition of the top row, which by definition is just $(f+g)_*$, and the composition of the curved arrows at the bottom, which is $f_*+g_*$, since $$f_* + g_* = (f_*\oplus g_*)\circ\Delta = (f\vee g)_* \circ (i_{1*}\oplus i_{2*}) \circ j \circ c_* = (f\vee g)_* \circ c_* = (f+g)_*\ .$$
Now we tackle this proof in three steps. First we show that the triangle on the right commutes. The isomorphism $i_{1*}\oplus i_{2*}: H_\bullet(\Sigma X)\oplus H_\bullet(\Sigma X)\to H_\bullet(\Sigma X\vee \Sigma X)$ is induced by the inclusion maps $i_1, i_2: \Sigma X \hookrightarrow \Sigma X\vee \Sigma X$. Since $(f\vee g)\circ i_1 = f$ and $(f\vee g)\circ i_2 = g$, using the universal property of direct sums, we can show that the composition $(f\vee g)_*\circ (i_{1*}\oplus i_{2*})$ is the direct sum map $f_*\oplus g_*$. (See below for the relevant diagram. The horizontal arrows at the bottom are the canonical injections into the direct sum.)
Next we want to construct the inverse of $i_{1*}\oplus i_{2*}$. Consider the maps $q_i: \Sigma X \vee \Sigma X \to \Sigma X, i = 1, 2$ that are identity on the $i$-th component, and constant on the other component. It's clear that $q_1\circ i_1$ and $q_2\circ i_2$ are the identity maps, and $q_1\circ i_2$ and $q_2 \circ i_1$ are constant maps, hence an easy calculation shows that $q_{1*}\times q_{2*}$ is the inverse of $i_{1*}\oplus i_{2*}$.
Finally to show that the left triangle commutes, note that $q_1\circ c$ and $q_2\circ c$ are both homotopic to the identity. (Think of this homotopy as sliding the equator to one of the poles.) This means that the induced maps $q_{1*}\circ c_*$ and $q_{2*} \circ c_*$ are equal to the identity, so $(q_{1*}\times q_{2*})\circ c_*$ is the diagonal map. (We can also draw a diagram for this proof. Even though it will seem a bit trivial, when we dualize it to prove the cohomology case later, it will become more useful.)
This completes the proof.
For the cohomology version of the statement, you essentially dualize all diagrams (reverse all arrows, direct sums become direct products and vice versa), and argue that in diagram 1 the top composition, which becomes $(f+g)^*$, is equal to the bottom composition, which becomes $f^* + g^*$. The central idea is similar, which is to break up $f^* + g^*$ into the composition of $f^*\times g^*$ with the addition $(x, y)\mapsto x+y$. The former is the curved arrow on the right, now pointing from $H^\bullet(Y)$ to $H^\bullet(\Sigma X)\times H^\bullet(\Sigma(X)$, and the latter is the curved arrow on the left, from $H^\bullet(\Sigma X)\times H^\bullet(\Sigma X)$ to $H^\bullet(\Sigma X)$. Commutativity on the left and right follows from the dualization of diagrams 2 and 3, respectively.
As an addendum, I want to relate this proof back to the diagram chasing in Hatcher. The commutativity of the diagrams that I'm proving here can be phrased as maps having equal values on all elements. So saying that $(f\vee g)_*\circ (i_{1*}\oplus i_{2*}) = f_*\oplus g_*$ is the same as saying for all $(x, y) \in H_\bullet(\Sigma X)\oplus H_\bullet(\Sigma X)$, we have $(f\vee g)_* (i_{1*}(x) + i_{2*}(y)) = f_*(x) + g_*(y)$. To prove this, you need only show it for the generators, which are of the form $(x, 0)$ and $(0, y)$. Similarly, to prove that $(q_{1*}\times q_{2*}) \circ c_* = \Delta$, you show that $(q_{1*}\times q_{2*})(c_*(x)) = (x, x)$ for all $x\in H_\bullet(\Sigma X)$. This is the route that Hatcher takes.