I’ll post a few common, trivial enough to not be written down anywhere, but not trivial enough for the average person to simply know after learning the basics of topology, theorems, mainly about algebra, topology, or algebraic topology.
Suppose a topological group G is T1. Then, G is Hausdorff.
Suppose the function f:GxG, (g, h)->gh^-1 is continuous. Then, clearly, the function i:G to G, h->h^-1 is continuous, as it is merely the restriction of the previous function to {e}xG, which is homeomorphic to G.
Naturally, the function GxG->G, (g, h)->gh is also continuous, as it is simply f(g, i(h)).
Clearly, if the identity element, e, is closed, then G is Hausdorff, as the preimage of {e} under f must also be closed, which is simply the diagonal {(g, g)|g in G}. Hence, G is Hausdorff iff the identity element {e} is closed.
The connected component H of {e} must be a closed normal subgroup; as:
Connected component is always closed
It’s a subgroup, because, given any g in H, gH=Hg=H. This is due to the fact x->gx is a homeomorphism, and g is an element of its image(take e->ge=g).
It’s normal, as the continuous image x->gxg^-1, x in H, must be connected, and maps e->geg^-1=e. Hence, it’s a subset of H.
Since H is a normal subgroup, we can define the group G/H, with the quotient topology(e.g. the function f(x)=xH=Hx, and G/H has the finest topology making this continuous)
If G is connected, so is G/H.
Proof:If G is connected, G=H.
If G is not connected, G/H isn’t either.
Proof:If T is a continuous non-constant function from G to {0, 1}, wlog assume T[H]=0. Then, the continuous function j from G/H to {0, 1} defined by j([e])=0 and j([x])=T(x) for x not in H is a continuous non-constant function, as it is the composition of the function mapping elements of G/H to their representatives in G, and T itself.
Clearly, G/H is Hausdorff, as the identity, {H}, is closed, because the quotient map is final with respect to projection to equivalence classes.
Suppose {H} is open. Then, G/H is discrete. Proof:Consider the quotient map p:G to G/H. Since G/H has the quotient topology, {e}=p(H) is open(closed) iff H is open(closed). Since H is a connected component, {e} is closed. If H is further open, {e} is clopen. Additionally, there is a homeomorphism from G/H to G/H by left translation by any particular element, so, if {e} is clopen, so is {g}, for any g in G/H, as x->gx is a homeomorphism, hence both an open and closed mapping.
From now on, let’s use H to denote the closure of the identity. Clearly, G/H is Hausdorff, and is connected iff G is connected.
Take any dense subset D of a (topological) group G. Denote it’s cardinality by |D|.
Then, given any continuous functions f, g:G to G/H, if f[D]=g[D] then f=g, since G/H is hausdorff. Hence, there is an injection from the set of continuous functions from G to G/H to the set of functions from D to G/H, (G/H)^D. In particular, this means there are at most |G/H|^|D| many continuous functions from G to G/H.
For example, take G=the real numbers and D=the rationals. Then, G/H=G=R. Hence, there are at most |R|^|Q|=(2N)^N=2^(N•N)=2^N continuous functions from R to itself. Hence, there are |R| many continuous functions from R to itself.
Clearly, G/H=G iff G is Hausdorff.
Theorem:If G is separable, there are at most |R| many continuous functions from G to G/H. The proof is nearly the exact same as the proof for the specific case of R.
Corollary:When G is a separable, uncountable Hausdorff group, there are exactly |R| many homeomorphisms from G to itself(assuming the continuum hypothesis). Hence, there is no separable Hausdorff group with strictly greater cardinality than R.
If G is compact, so is G/H, as it is the image of a compact space under a continuous function.
Multiplication is continuous. Hence, given any two loops p, q in G, and the fact t->(p(t), q(t)) is a continuous function into GxG, pointwise multiplication of paths is continuous.
