Let $X$ be a differentiable manifold. Consider the homomorphism: $\phi: H^n(X,\mathbb{Z})\to H^n(X,\mathbb{R})$ induced by the inclusion of the constant sheaves $\mathbb{Z}\subset\mathbb{R}$. Let $\tilde{H}^n(X,\mathbb{Z})$ denote the image of ${H}^n(X,\mathbb{Z})$ in $H^n(X,\mathbb{R})$.
My question is: how to show $\tilde{H}^n(X,\mathbb{Z})$ is integral cohomology modulo torsion?
As mentioned in the comment by @LordSharktheUnknown, the universal coefficient theorem (applied to both $\mathbb Z$ and $\mathbb R$) implies that $\tilde H^k(M,\mathbb Z)$ coincides with the image of $Hom_{\mathbb Z}(H_k(M,\mathbb Z),\mathbb Z)$ in $Hom_{\mathbb Z}(H_k(M,\mathbb Z),\mathbb R)$ under the map induced by the inclusion of $\mathbb Z$ into $\mathbb R$. Now any homomorphism from an abelian group to $\mathbb R$ vanishes on the torsion subgroup. Restricting to the free part, you get the inclusion of $\mathbb Z^A$ into $\mathbb R^A$, where $A$ is a set of generators of the free part. So the image is a lattice in the vector space $H^k(M,\mathbb R)$ (but certainly not a linear subspace). In many cases of interest, $H^*(M,\mathbb R)$ is finite dimensional, then this is just like as specific embedding of $\mathbb Z^N$ into $\mathbb R^N$.