Likelihood is not "proportional to" a single probability density? In various places it says that the likelihood (e.g. in the Bayes formula) is "proportional to a probablility".
For example
https://alexanderetz.com/2015/04/15/understanding-bayes-a-look-at-the-likelihood/
"likelihood is proportional to a probability"
A similar statement is in the book Think Bayes,
"likelihood doesn't need to compute a probability, it only has to compute something proportional to probability"
Lastly, in a crossvalidated discussion 
https://stats.stackexchange.com/questions/2641/what­is­the­difference­between­likelihood­and­%20probability
there is a more specific statement "the likelihood function is proportional to the probability of the observed data."  I am not sure if that is correct or not.
Notation, $P(A|B) = P(B|A) P(A) / P(B)$.  The likelihood is $P(B|A)$.
The question: If I am understanding however, the likelihood is proportional to a different probability
for each value of $A$. That is, there is no single constant $c$ such that the likelihood
is equal to some version of the probability $P(B)$, i.e. this statement is false
$$
    P(B) = c \cdot P(B|A)
$$
Rather, it would have to be
$$
    P(B) = c(A) \cdot P(B|A)
$$
meaning that the constant of proportionality $c$ varies with each choice of $A$.
Is this true?  If so, it seems to me that saying likelihood is proportional to a probability is vacuous.
You can always make something "proportional to" something else if the proportionality is a function
rather than a constant.
This question is somewhat related to these previous questions about likelihood, which I have read, however I believe this is a different question (albeit maybe one that could be "derived from" the answers to other questions, buy someone smarter than me!)
What does "likelihood is only defined up to a multiplicative constant of proportionality" mean in practice?
When is likelihood also a probability distribution?
EDIT: Maybe another way of asking the question is this: if the likelihood is proportional to a probability, which probability is it proportional to?  Is it the probability $P(B|A)$ regarded as a function of B with A held fixed, or is it the marginal probability P(B), or is it some other (generic) probability?
EDIT: here is an attempt at an example:
A Normal distribution is parameterized by mean and variance. Let's pick something simpler,a PMF that is parameterized only by a single parameter $M$.  It assigns probabilities to integer values, 
and the parameter translates the PMF on the integer axis.
Here is the PMF for the parameter value M=5:
$$
\begin{align*}
&  P(X=5|M=5) = .5  \\
&  P(X=6|M=5) = .3  \\
&  P(X=7|M=5) = .2  \\
\end{align*}
$$
And the PMF for parameter value M=6:
$$
\begin{align*}
&  P(X=5|M=6) = 0  \\
&  P(X=6|M=6) = .5 \\
&  P(X=7|M=6) = .3  \\
&  P(X=8|M=6) = .2  \\
\end{align*}
$$
More generically, the PMF is like P(X=M)=0.5, P(X=M+1)=.3, P(X=M+2)=.2, zero otherwise.
Now consider the likelihood form of this, where the data is given as the fixed value X=5
and the parameter $M$ is what varies:
$$
\begin{align*}
&  P(X=5|M=3) = .2  \\
&  P(X=5|M=4) = .3  \\
&  P(X=5|M=5) = .5  \\
\end{align*}
$$
So in the PMF case the shape is (.5,.3,.2),  whereas in the likelihood case the shape is (.2,.3,.5).  There is no single constant that can make these equal.
What is my mistake?
 A: The likelihood function is defined in the context of parametric distribution. Given a family of pmf's $p_θ(x)$, $x\in\mathfrak{X}$ meaning that, for every and all values of $θ$, $p_θ(\cdot)$ is a probability density over $\mathfrak{X}$ [wrt to a well-defined dominating measure] and an observation $x^o$, supposedly generated from one of these pmf's, ${p_{θ^o}}(\cdot)$ say, the likelihood function$$ℓ:\Theta \longrightarrow \mathbb{R}^+$$is a function (of $θ$) proportional to $p_θ(x^o)$, 
$$\ell(θ) \propto p_θ(x^o)$$
proportional in the sense that there exists a constant $κ$ such that $$ℓ(θ)=κp_θ(x^o)$$ $κ$ being constant in $θ$ (but possibly depending on $x^o$, dependence that does not matter since $x^o$ is observed, hence fixed). 

If one does not adopt a Bayesian or a fiducial approach, the
  likelihood cannot be defined more precisely, that is, there is no
  general principle towards selecting a value of $κ$. Hence the statement that it is proportional to the pmf, taken at $x^o$, a probability density in $x$ and not in $\theta$. The likelihood is not a probability distribution in $\theta$ as the "likely" in "likelihood" refers to the probability of observing the observed $x^o$ given the inputed value of the parameter $\theta$.

A: Suppose three biased coins have probabilities $0.6,\,\,0.7,\,\, 0.8$ of "heads".
Suppose you are equally likely to have any of these three coins in your hand. You toss it and it turns up "heads". Then you have
$$
\begin{array}{cccc}
\text{prior} & \text{likelihood} & \text{product} & \text{posterior} \\
\downarrow & \downarrow & \downarrow & \downarrow \\
1/3 & 0.6 & 0.6/3 & 6/(6+7+8) = 6/21 \\
1/3 & 0.7 & 0.7/3 & 7/(6+7+8) = 7/21 \\
1/3 & 0.8 & 0.8/3 & 8/(6+7+8) = 8/21
\end{array}
$$
Multiplication of the "product" column by the normalizing constant $3/(6+7+8)$ yields the "posterior" column.
The "likelihood" column is not proportional to the "posterior" column in cases where the prior probabilities are not all equal to each other.
