Show there is no map $f: S^1\vee S^1 \vee S^2 \to \mathbb T$ that induces isomorphisms for every homology group I get that the wedge sum of two 1-spheres and a 2-sphere has the same homology groups as a torus, but how would I show there is no map that induces an isomorphism between each homology group? I assume it is something to do with where the generators would be sent but I am not so sure.
 A: Denote by $[X,Y]$ the set of homotopy classes of maps between $X$ and $Y$. Then
$$[S^2,\mathbb{T}]=[S^2,S^1\times S^1]\equiv [S^2,S^1]\times [S^2,S^1]=\{*\}$$
because all maps $S^2\to S^1$ are homotopic (because $\mathbb{R}$ is the universal cover of $S^1$).
This shows that all maps $S^2\to \mathbb{T}$ are homotopic. In particular every such map induces the zero morphism on homology groups.
Now given $f:S^1\vee S^1\vee S^2\to \mathbb{T}$ act with $H_2$ on it. Then we have
$$H_2(f):H_2(S^1\vee S^1\vee S^2)\to H_2(\mathbb{T})$$
It is well known that homology of wedge sum is direct product of homologies (in a natural way), and so we can rewrite the above as
$$H_2(f):H_2(S^1)\times H_2(S^1)\times H_2(S^2)\to H_2(\mathbb{T})$$
$H_2(S^1)$ is trivial, while $f$ restricted to $S_2$ is null homotopic, and so it induces zero on the $H_2(S^2)$. Therefore $H_2(f)$ is the zero map, and since $H_2(\mathbb{T})$ is not zero, then it cannot be an isomorphism.
A: There is another way to do this problem, namely by showing that for the space $X = S^n \times S^m$ (take $n,m \neq 0$ to avoid a trivial problem), there is no map $f:A \vee B \to X$ inducing an isomorphism on homology groups for any based $A$ and $B$, unless one of $A$ and $B$ is the one-point space and the other is $X$.
To see this, first note the following necessary condition. $\tilde{H}_k(X)$ is $\mathbb{Z}$ in degrees $n, m, n+m$ and trivial otherwise (if $n=m$ then take a $\mathbb{Z}^2$ in degree $n$), so suppose the same is true for $\tilde{H}_k(A \vee B) = \tilde{H}_k(A) \oplus \tilde{H}_k(B)$. $A$ and $B$ must both be path-connected, and neither can have trivial homology since otherwise the one with trivial homology is contractible. So we must be in one of two cases, since the problem is symmetric in $A$ and $B$ and $n$ and $m$: (1) $A$ has the homology in degrees $n$ and $m$ and $B$ has the homology in degree $n+m$, or (2) $A$ has homology in degrees $n$ and $n+m$ and $B$ has the homology in degree $m$.
Now consider the cohomology rings. One has the ring isomorphism $i_A^* \times i_B^*: \tilde{H}^*(A \vee B) \to \tilde{H}^*(A) \times \tilde{H}^*(B)$ where $i_A, i_B$ are inclusions of $A$ and $B$ into the wedge sum, so suppose that such an $f$ exists and consider the composite $(i_A^* \times i_B^*) \circ f^*: \tilde{H}^*X \to \tilde{H}^*(A) \times \tilde{H}^*(B)$. If $f$ induces an isomorphism on cohomology groups, then it induces an isomorphism on cohomology rings, so this composite must also be a ring isomorphism.
In $X$, for $u$ the generator of $\tilde{H}^n(S^n)$ and $v$ the generator of $\tilde{H}^m(S^m)$ we have $\tilde{H}^{n+m}(X)$ generated by their cup product $uv$ by the Kunneth theorem. Thus the product of a generator for $\tilde{H}^n(A \vee B)$ and $\tilde{H}^m(A \vee B)$ must generate $\tilde{H}^{n+m}(A \vee B)$.
In case (1), both these generators come from $\tilde{H}^*(A)$, so their product must lie in $\tilde{H}^{n+m}(A)$, which is zero unless either $n$ or $m$ is zero. In case (2), one of these generators comes from $\tilde{H}^n(A)$ and the other from $\tilde{H}^m(B)$, but the product of these classes is zero in $\tilde{H}^*(A) \times \tilde{H}^*(B)$.
Thus it is impossible for $(i_A^* \times i_B^*) \circ f^*$ to be an isomorphism of cohomology rings, hence it is impossible for $f^*$. As a consequence, this shows that a product of spheres never splits as a nontrivial wedge sum.
