Is every  group object in TopMan a Lie group? Recall that a Lie group is a group object in the category of C∞ manifolds.
If I have a group object in the category of topological manifolds, can I necessarily equip it with a smooth structure so that all the group operations are smooth?  If so, how unique is this structure?
Is a continuous group-homomorphism between two Lie groups necessarily smooth?
 A: I just wanted to add that there is a fairly easy proof for your final question:  Is every continuous homomorphism between Lie groups actually smooth?
The theorem we need is the closed subgroup theorem (also called the Cartan Theorem): If H is a topologically closed subgroup of a Lie group G, then H is actually an embedded Lie subgroup.
Granting this, one proves all continuous homomorphisms are smooth as follows:
Given Lie groups H and G with $f:H\rightarrow G$ a continuous homomorphism, consider the subgroup $K$ of $H\times G$ given by the graph of $f$.  The graph is a closed subset of $H\times G$ precisely because $f$ is continuous, and hence, by the closed subgroup theorem, the graph is an embedded smooth submanifold of $H\times G$.  Thus, the restriction of the two canonical projection maps $\pi_1:H\times G\rightarrow H$ and $\pi_2:H\times G\rightarrow G$ are smooth when restricted to K.
Now, $\pi_1$ restricted to $K$ is clearly* a diffeomorphism onto $H$, and hence has a smooth inverse and so is smooth.  But then we find that $f = \pi_2\circ \pi_1^{-1}$ is a composition of smooth maps, and hence is smooth.  (To be clearer, the $\pi_1^{-1}$ means the inverse of $\pi_1:K\rightarrow H$.)
*- (Edited in due to comments).  One knows by Sard's theorem that there is a point $p\in K$ such that $d_p \pi_1$ is invertible (of full rank).  I claim that this implies that for all $q\in K$, $d_q \pi_1$ is invertible.  The point is that $\pi_1$ is group homomorphism, which is the same as saying $\pi_1\circ L_{qp^{-1}} = L_{\pi_1(qp^{-1})}\circ \pi_1$, where $L_g$ denotes left multiplication by $g: L_g(h) = gh$.  Taking the differentials at p on each side of this equation and using the chain rule, one finds 
$$d_q \pi_1 \circ d_p L_{qp^{-1}} = d_{\pi_1(p)}L_{\pi_1(qp^{-1})}\circ d_p \pi_1.$$
The fact that $L_g$ is a diffeomorphism (with inverse $L_{g^{-1}}$) implies that $dL$ is invertible at any point, and hence we see that
$$d_q\pi_1 = d_{\pi_1(p)}L_{\pi_1{qp^{-1}}}\circ d_p \pi_1\circ d_pL_{pq^{-1}};$$ i.e., that $d_q \pi_1$ is a composition of invertible maps, and hence is itself invertible.
A: As Greg Kuperberg indicated in the comments, this is Hilbert's 5th Problem.  The answer is yes, a theorem of Gleason, Montgomery and Zippin from the 1950's.
