What is Hom$(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})$? I am trying to compute the singular cohomology of $\mathbb{N}\subset\mathbb{R}$. I have that $C_0(\mathbb{N})\cong C_1(\mathbb{N})\cong\mathbb{Z}^{\oplus\mathbb{N}}$, where I mean $\mathbb{Z}^{\oplus\mathbb{N}}=\{(n_0,n_1,n_2,...)|n_i\in\mathbb{Z},n_i=0\text{ for all but finitely many }i\}$. Moreover, I know that $d_1=d_0=0$.
I know that that Hom$(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})\cong C^0(\mathbb{N})\cong C^1(\mathbb{N})$.
I am fairly sure that Hom$(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})\cong \mathbb{Z}^\mathbb{N}$, where $\mathbb{Z}^\mathbb{N}=\{(n_0,n_1,n_2,...|n_i\in\mathbb{Z}\}$. I am, however, unsure how to prove this.
 A: This gives the same answer as the one posted, but invoking categorical arguments.
Since the direct sum is the coproduct, and maps from a coproduct correspond to families of maps from the constituents, each morphism $f\colon \oplus_{i\in I}A_i\to B$ corresponds to a family of morphism $\{f_i\colon A_i\to B\}_{i\in I}$. That is, there is a natural isomorphism (induced by the universal property of the direct sum)
$$\mathrm{Hom}(\oplus_{i\in I}A_i,B)\cong \prod_{i\in I}\mathrm{Hom}(A_i,B).$$
(Similarly, maps into the product correspond to families of maps into the factors).
For $A_i=B=\mathbb{Z}$, each $\mathrm{Hom}(A_i,B)\cong\mathbb{Z}$, which gives
$$\mathrm{Hom}(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})\cong \prod_{i\in\mathbb{N}}\mathbb{Z}= \mathbb{Z}^{\mathbb{N}}.$$
A: Let $x\in\mathbb{Z}^\mathbb{N}, x=(x_0,x_1,x_2,...)$. Let $f_x:\mathbb{Z}^{\oplus\mathbb{N}}\rightarrow\mathbb{Z};f_x(y)=x_0y_0+x_1y_1+x_2y_2+...$
Then $f_x\in\text{Hom}(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})$, as only finitely many $y_i$ are nonzero for any $y\in\mathbb{Z}^{\oplus\mathbb{N}}$.
Moreover, if $f_x=0$, then, letting $y^i\in\mathbb{Z}^{\oplus\mathbb{N}}$ be the element with $y^i_j=0\ \forall j\neq i,y^i_i=1$, we have $f_x(y^i)=x_i=0$ for every $i\in\mathbb{N}$. Thus $x=0$. Furthermore, any $f\in\text{Hom}(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})$ is totally determined by the values it takes on the $y^i$s, so if $x=(f(y^0),f(y^1),f(y^2),...)$, then $f_x=f$.
Thus the homomorphism $\phi:\mathbb{Z}^\mathbb{N}\rightarrow\text{Hom}(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z});x\mapsto f_x$ is an isomorphism.
