Does $\mathbf{T} \vdash \phi \; \iff \; \mathbf{M} \models \phi$ equivalence hold? I re-read through my textbook section about the completeness theorem, and it states:

For a countable signature $\mathcal{L}$, and a consistent set of sentences (closed-formulae) $\mathbf{T}$, there exists a model for said consistent set of sentences. It is also true that if $\lnot(\mathcal{T} \vdash \phi)$ (for some $\mathcal{L}$-formula $\phi$) then $\mathbf{T} + \lnot \phi$ has a model.

Now, my question is, that for some consistent theory, shouldn't
$$\mathbf{T} \vdash \phi \; \iff \; \mathbf{M} \models \phi$$
For $\mathbf{M} \models \mathbf{T}$ hold? I tried to take the contrapositive for $\mathbf{M} \models \phi \; \to \; \mathbf{T} \vdash \phi$, and say that $\lnot (\mathbf{T} \vdash \phi) \to \lnot (\mathbf{M} \models \phi)$ to disprove myself but it still leaves me with the gap of "why should there be no model?" or is that we can't know (thus that statement isn't always true for consistent theories, the equivalence that is)? Or am I just not asking the right question here, or does that en fact disprove it?
Not to be confused with: "every theory is consistent if it has a model" (which is a consequence of completeness theorem [or a re-formulation of the theorem]).

Tl;dr Does the soundness theorem go both ways?
 A: Mark Kamsma's comment is the crux of the issue here – there are two ways of interpreting your question:

*

*Is it the case that $\mathbf{T}\vdash\phi$ if and only if $\mathbf{M}\models\phi$ for every model $\mathbf{M}\models\mathbf{T}$?

*Is it the case that $\mathbf{T}\vdash\phi$ if and only if $\mathbf{M}\models\phi$ for some model $\mathbf{M}\models\mathbf{T}$?

Note that the "only if" direction follows for both statements by the soundness theorem. So the question is really which of the "if" implications, if either, holds. The answer is that the "if" implication in Question 1 does hold, and that the "if" implication in Question 2 does not hold.
Why does it hold in the first case? Well, your idea of showing it by contrapositive is a good one; suppose that $\neg(\mathbf{T}\vdash\phi)$. The negation of the statement "$\mathbf{M}\models\phi$ for every $\mathbf{M}\models\mathbf{T}$" is the statement "$\neg(\mathbf{M}\models\phi)$ for some model $\mathbf{M}\models\mathbf{T}$", so to get the desired result we need only find a single model of $\mathbf{T}$ that does not model $\phi$. We can do this using the paragraph you quote from your textbook, which tells us there exists a model $\mathbf{M}$ of the theory $\mathbf{T}+\neg\phi$. Then $\mathbf{M}\models\neg\phi$, which is the same thing as $\neg(\mathbf{M}\models\phi)$. (Why?) Since also $\mathbf{M}\models\mathbf{T}$, this gives us a model of $\mathbf{T}$ that does not model $\phi$, precisely as desired.
Why does it fail to hold in the second case? Well, there are many examples of this; the simplest is arguably to take $\mathcal{L}$ to be the "empty language" and $\mathbf{T}$ to be the "empty theory", ie to take $\mathcal{L}=\varnothing$ and $\mathbf{T}=\varnothing$. Then any set, with no additional structure at all, is a model of $\mathbf{T}$! Now suppose that $\phi$ is the sentence $\exists x\forall y(x=y)$; can you find a model $\mathbf{M}\models T$, ie a set $\mathbf{M}$, such that $\mathbf{M}\models\phi$? If the implication in Question 2 held, then this would mean that $\mathbf{T}\vdash \phi$, and by the soundness theorem this would mean that $\mathbf{N}\models\phi$ for every model $\mathbf{N}\models \mathbf{T}$. But recall that a model of $\mathbf{T}$ is just a pure set, so for instance the structure $\mathbf{N}=\{\star,\bullet\}$ is a model of $\mathbf{T}$; can you see why $\neg(\mathbf{N}\models \phi$)? So this cannot be the case, and this gives the desired counterexample to the "if" implication in Question 2.
If that example feels too vacuous or abstract, a more "natural" example can be found in Troposphere's comment below your question. But the main takeaway is that the statement is Question 1 is correct, and the statement is Question 2 is false.

A few additional remarks:

*

*As ℋolo comments below, one normally abbreviates the statement "$\mathbf{M}\models\phi$ for every model $\mathbf{M}\models\mathbf{T}$" by merely writing $\mathbf{T}\models\phi$. So the statement in Question 1 above can be rewritten as the pleasing symmetry that $\mathbf{T}\vdash\phi$ if and only if $\mathbf{T}\models\phi$!


*As Troposphere comments below, the formulation by ℋolo is actually much more natural than the usual slogan for the completeness theorem, which is that "every consistent theory has a model". One can argue that consistency is actually of secondary interest to satisfiability, and that what we really care about in logic is the models of the theory, rather than one or another proof system associated to it. (See Joel David Hamkins' answer here for an argument along those lines.) So, as Troposphere points out, from that point of view the significance of the completeness theorem is that it says our syntatic system is "strong enough" to successfully capture the meaning of truth in our semantic system (which is perhaps the thing we really care about), rather than saying that our semantic system is "weak enough" to sucessfully capture the meaning of truth in our syntatic system. See Troposphere's second comment below for a very nice example where the equivalence of these two ideas fails.


*As Alex comments below, there is indeed a third way of framing your question, which is whether, for a fixed $\mathbf{T}$, $\mathbf{M}\models\mathbf{T}$, and $\phi$, we have that $\mathbf{T}\vdash\phi$ if and only if $\mathbf{M}\models \phi$. You can find counterexamples to this exactly as you can find counterexamples to the statement in Question 2 above; either consider the case $\mathbf{T}=\mathcal{L}=\varnothing$ as I do, or consider Troposphere's example in the comments to your question. Of course, it will certainly be possible to find examples of $\mathbf{T}$ and $\mathbf{M}$ such that $\neg(\mathbf{T}\vdash\phi)$ and $\neg(\mathbf{M}\models \phi)$ too! (In fact, that is precisely the content of the statement in Question 1 above.) But the point is that this will not happen in general. However, this framing of the question poses something different to think about; if there does exist a model $\mathbf{M}\models\mathbf{T}$ such that $\mathbf{M}\models\phi$ implies $\mathbf{T}\vdash\phi$ for every $\phi$, then what can we say about $\mathbf{T}$? In turns out that, if a model of $\mathbf{T}$ exists with that property, then actually every model of $\mathbf{T}$ will have that property; see if you can prove this as an exercise using the soundness theorem. In this case we say that $\mathbf{T}$ is a complete theory; complete theories are of great interest in much of modern model theory.
