Scars are a result of the natural healing process that occurs when the skin repairs itself after wounds, trauma, burns, surgical incision, or disease. Normal skin tissue is replaced with scar tissue to close open wounds and prevent infection. Scars can be painful, cause itching, and limit mobility. Many scars are primarily a cosmetic concern, but their presence may have a significant negative i...

Solution. Let α : Z/nZ → Z/mnZ be defined by α(1) = m. Since 〈1〉 = Z/nZ, this determines the entire homomorphism. Let x ∈ kerα. Then mx ≡ 0, which implies mn | mx so n | x. But since 0 ≤ x < n, we must have x = 0. Therefore the kernel of α is trivial, so α is an injection. Since m | mn, by the fundamental theorem on cyclic groups Z/mnZ has a subgroup of order m isomorphic to Z/mZ. Let β : Z/mnZ...

t(x · 1 + 0 · s) = t · x = 0. Therefore f ′ is injective. Now we need to show that im f ′ = ker g′. We have g′(f ′(x/s)) = g(f(x)/s) = (g ◦ f)(x)/s = 0, so im f ′ ⊂ ker g′. Conversely, if g′(y/s) = 0, then g(y)/s = 0, so there exists t ∈ S so that t · g(y) = g(t · y) = 0. Therefore t · y ∈ ker g, so t · y ∈ im f . Let x ∈ X so that f(x) = t · y. Then f ′(x/(st)) = f(x)/(st) = (t · y)/(st) = y/s...

Hypoxia is a central component of the tumor microenvironment and represents a major source of therapeutic failure in cancer therapy. Recent work has provided a wealth of evidence that noncoding RNAs and, in particular, microRNAs, are significant members of the adaptive response to low oxygen in tumors. All published studies agree that miR-210 specifically is a robust target of hypoxia-inducible...

Boeing has been working on the development and implementation of STEP AP 210 since its inception. Using a combination of internally funded activities, external government contracts and multi-company pilots, Boeing has developed prototype tools to utilize AP 210 and is poised to move these tools into production. This paper begins with a brief overview of STEP and AP 210. The status of available ...

We claim that monomorphisms in Sets and Groups are the usual injective maps and homomorphisms. First, suppose that f : A → B is a set injection and that fg = gh. Then for all c ∈ C, we have f(g(c)) = f(h(c)) implies g(c) = h(c) by injectivitiy. Hence g = h. Conversely, suppose f : A → B is not an injection. Then let f(a) = f(a′) for some a 6= a′ ∈ A. Then let g(c) = a and h(c) = a′ for all c ∈ ...