Here $C(f)=Cone(f:Xto Y)$ is the pushout of

$require{AMScd}$

begin{CD}

X @>>>C(X)\

@V{f}VV @VVV\

Y @>>> C(f)

end{CD}

And $C(X):=frac{Xtimes(0,1)}{Xtimes 1}$ denotes the cone of $X$.

For notational convenience we denote $f:S^5_ato S^5_b$ and $g:S^5_bto S^5_c$. Then we know

$$C(gcirc f)=frac{S^5_csqcup C(S^5_a)}{gcirc f(x)sim ((x,0))}$$

First we define the map $displaystyle{phi:S^5_csqcup C(S^5_a)to C(g)=frac{S^5_csqcup C(S^5_b)}{g(x)sim ((x,0))}}$ by

$$phi(x)=(x) forall xin S^5_c$$ and $$phi((x,t))=(f(x),t) forall ((x,t))in C(S^5_a)$$

Then $phi(gcirc f(x))=((f(x),0))=phi(((x,0)))$. Hence, by the universal property of quotient spaces $phi$ induces the map

$$psi:C(gcirc f)to C(g)$$

Now I have to calvculate the $psi_*:H_k(C(gcirc f))to H_k(C(g))$.

I know that as $f,g$ has degree $2$, so $gcirc f$ has $4$. Hence, $$C(gcirc f)cong M(Bbb{Z}/4,5)text{ and } C(g)cong M(Bbb{Z}/2,5)$$

where $H_5(M(Bbb{Z}/n,k))=Bbb{Z}/n, H_0(M(Bbb{Z}/n,k))=Bbb{Z}$ and $0$ otherwise.

Can anyone help me to compute $psi_*:H_k(C(gcirc f))to H_k(C(g))$ for $k=0,5$? Thanks for help in advance.