
10 Madhu Ram - Bijan Davvaz
Proof. Since M is a vector subspace of L, it is a super quasi-topological vector space with the
topology induced by the topology of L. Let ϕ and π be the canonical mappings from M to M/N
and from L to L/N, respectively. Let U be open for the quotient topology of M/N. Then ϕ
−1
(U)
is open in M, so ϕ
−1
(U) = M ∩ V where V is an open subset of L.
Claim: U = (M/N) ∩ π(V ).
Let η ∈ (M/N) ∩ π(V ). Then η = x + N for some x ∈ M and η = v + N for some v ∈ V .
This implies that v − x ∈ N, so v ∈ x + N ⊆ M + N = M . Therefore, v ∈ M ∩ V = ϕ
−1
(U), so
η = v + N ∈ U . Clearly, U ⊆ (M/N) ∩ π(V ) and the claim follows.
Now let A be open in M/N for the topology on M/N induced by the quotient topology of
L/N. Then A = (M/N) ∩B for some open subset B of L/N. Obviously, ϕ
−1
(A) = M ∩π
−1
(B)
is an open subset of M . This means that A is open for the quotient topology of M/N.
Corollary 4.1. If M and N are vector subspaces of a super quasi-topological vector space L, then
the quotient topology on (M + N)/N is identical with the topology on it induced by the quotient
topology of L/N.
Theorem 4.6. Let f be a linear map from a super quasi-topological vector space L to a super
quasi-topological vector space E, and let M be a vector subspace of L that is contained in the
kernel of f. The linear map g from L/M to E satisfying g ◦ π = f is continuous (open) if and
only if f is continuous (open).
Proof. The necessity part follows from Theorem 4.1. Conversely, assume f is continuous. Let U
be a neighborhood of 0 in E. Then g
−1
(U) = π ◦ f
−1
(U), so g is continuous at 0. By Theorem
3.14, g is continuous.
Theorem 4.7. If M is a vector subspace of a super quasi-topological vector space L, and if M
and L/M are both Hausdorff, then L is Hausdorff.
Proof. Let x be an element of L such that x 6= 0 and let x ∈ U for each U ∈ V, the neighborhood
filter of 0 in L. Since M is Hausdorff, x /∈ M . Then x + M and M are two distinct elements of
L/M. As L/M is Hausdorff, there are disjoint open sets A and B for the quotient topology of
L/M containing x + M and M , respectively. By Theorem 3.14, π
−1
(A) is a neighborhood of x
and π
−1
(B) is a neighborhood of 0 in L. By assumption, x ∈ π
−1
(B), so x ∈ π
−1
(A) ∩ π
−1
(B),
a contradiction. By Theorem 3.9, L is Hausdorff.
Theorem 4.8. If M is the connected component of zero in a super quasi-topological vector space
L, and M a vector subspace, then L/M is totally disconnected.
Proof. Let K be a closed subset of L/M such that π
−1
(K) is disconnected. We will show that K
is disconnected. Let A and B be non-empty subsets of π
−1
(K) such that A ∪ B = π
−1
(K) and
A ∩ B = ∅. As for each x ∈ A, x + M is connected subset of π
−1
(K) and hence A = A + M =
π
−1
(π(A)).
Similarly, B = π
−1
(π(B)).
Since π(A) ∩ π(B) = π(A ∩ B) = ∅ and (L/M )\π(A) = π(L\A) which is open, so π(A) is
closed subset of L/M. Similarly, π(B) is closed in L/M. As
π(A) ∪π(B) = π(A ∪ B) = π(π
−1
(K)) = K,
so K is disconnected. Now,
if C is the connected component of zero in L/M , and if there is a point π(x) of L/M such
that π(x) ∈ C and x /∈ M , then π
−1
(C) would be disconnected, which is a contradiction. It ends
the proof.
Divulgaciones Matem´aticas Vol. 23-24, No. 1-2 (2022-2023), pp. 1–11